Mathematical Analysis - Hong Kong Baptist Universityxwu/m1111/MathAnal.pdf8 Mathematical Analysis. Chapter 1 Sets and Numbers 1.1 Sets Set theory is the foundation in which virtually

Mathematical Analysis

Hongci Huang and Xiaonan Wu

Department of Mathematics

Hong Kong Baptist University

August 18, 2015

2

Contents

1 Sets and Numbers 9

1.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.2 Real Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3 Inequalities and Algebraic Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.4 Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Limit Theorey 21

2.1 Convergence and Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.1.1 Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.1.2 Convergence and Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.1.3 Uniqueness of Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.1.4 Boundedness of Convergent Sequence . . . . . . . . . . . . . . . . . . . . . . . . 26

2.1.5 Operations on Convergent Sequences . . . . . . . . . . . . . . . . . . . . . . . . 28

2.1.6 Squeezing Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.2 Some Important Theorems in Limit Theory . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.2.1 Dedekind Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.2.2 Least Upper Bound Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.2.3 Monotone Convergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.2.4 Nested Interval Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.2.5 Heine-Borel Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.2.6 Bolzano-Weierstrass Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.2.7 Cauchy Convergence Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.3 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3

4

3 Continuity 51

3.1 Limit of Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.1.1 Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.1.2 Operations and properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

3.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3.2.2 Operations and composition on continuous functions . . . . . . . . . . . . . . . . 57

3.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

3.3 Extreme Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3.3.1 Maximizer and minimizer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3.3.2 Extreme value theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3.3.3 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.4 Intermediate Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.5 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.6 Inverse Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.7 Continuity of Elementary Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

3.7.1 Power Function with Rational Exponential . . . . . . . . . . . . . . . . . . . . . 66

3.7.2 Exponential Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

3.7.3 Logarithmic Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

3.7.4 Power Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

3.8 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

4 Differentiation 79

4.1 Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4.1.2 Physical and geometric interpretation . . . . . . . . . . . . . . . . . . . . . . . . 81

4.1.3 Algebra of Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

4.2 Differentiating Inverses and Compositions . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.2.1 Inverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.2.2 Composition and Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

4.3 Derivatives of Elementary Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

4.4 Mean Value Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

4.5 High (Order) Derivatives and Differential . . . . . . . . . . . . . . . . . . . . . . . . . . 88

4.6 Applications of Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

4.6.1 L’Hopital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5

4.6.2 Extreme values of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

4.6.3 Concavity and graph of functions . . . . . . . . . . . . . . . . . . . . . . . . . . 94

4.7 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

5 Indefinite Integral 103

5.1 Definition and Some Basic Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

5.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

5.2 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

5.3 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

5.4 Integration of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

5.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

6 Integration 113

6.1 Darboux Sums and Definition of Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

6.1.1 Darboux Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

6.1.2 Refinement Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

6.1.3 Definition of Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

6.1.4 Interpretation in Physics and Geometry . . . . . . . . . . . . . . . . . . . . . . . 116

6.2 Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

6.2.1 Integrability Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

6.2.2 Two Kinds of Integrable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 118

6.2.3 Convergence of Darboux Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

6.2.4 Convergence of Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

6.3 Linearity, Monotonicity and Additivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

6.3.1 Linearity of Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

6.3.2 Monotonicity of Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

6.3.3 Additivity over Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

6.4 Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

6.4.1 First Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . 127

6.4.2 Mean Value Theorem for Integral . . . . . . . . . . . . . . . . . . . . . . . . . . 129

6.4.3 Second Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . 130

6.4.4 Calculation of Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

6.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

7 Series of Numbers 141

7.1 Definition and Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

6

7.2 Series with Nonnegative Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

7.3 Convergence of General Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

7.4 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

8 Series of Functions 153

8.1 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

8.1.1 Pointwise Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

8.1.2 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

8.1.3 Test for Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

8.2 Uniform Limit of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

8.2.1 Uniform Limit of Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . 160

8.2.2 Uniform Limit of Integrable Functions . . . . . . . . . . . . . . . . . . . . . . . . 160

8.2.3 Uniform Limit of Differentiable Functions . . . . . . . . . . . . . . . . . . . . . 162

8.3 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

8.3.1 Radius of Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

8.3.2 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

8.4 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

8.4.1 Taylor Polynomials and Order of Contact . . . . . . . . . . . . . . . . . . . . . . 169

8.4.2 Convergence of Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

8.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

9 Fourier Series 181

9.1 Fourier Series and Fourier Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

9.2 Convergence of Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

9.3 Integration and Differentiation of Fourier Series . . . . . . . . . . . . . . . . . . . . . . . 189

9.4 Fourier Expansion on Finite Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

9.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

10 Improper Integral 195

10.1 Integration on an Unbounded Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

10.1.1 Nonnegative Integrand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

10.1.2 Absolute Convergence Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

10.1.3 Improper Integral on (−∞, b] and (−∞,+∞) . . . . . . . . . . . . . . . . . . . 198

10.2 Integral with an Unbounded Integrand . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

10.2.1 Nonnegative Integrand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

10.2.2 Absolute Convergence Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

Chapter 1: Sets and Numbers 7

10.2.3 Other Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

10.3 Gamma Function and Beta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

10.3.1 Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

10.3.2 Beta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

10.4 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

8 Mathematical Analysis

Chapter 1

Sets and Numbers

1.1 Sets

Set theory is the foundation in which virtually all mathematics is constructed. Set theory is quite easy

to learn and use.

Set. A collection of well-defined objects is a set. ”object” is also called ”element” or ”member”. ”Well-

defined” means that an object is distinguishable by some test whether it is in the set or not. The sets

discussed in our course are mainly real number sets.

Notation: Capital letters A,B,C are usually used to denote Sets, and small letters a, b, c, x, y, z are used

to denote elements.

• x ∈ A means element x in A.

• x �∈ A (or x∈A) means x not in A.

• A = {x | conditions on x}.

Examples.

1. A = {x | x2 + 3x+ 1 = 0}⇔ A =

{−3+

√5

2 , −3−√5

2

}2. A = {n ∈ N | n can be divided by 3} ⇔ A = {3, 6, 9, · · · } or A = {n | n = 3k, k ∈ N}, where N

is the set of natural numbers.

Empty set. A set that has no members, denoted by ∅.

A = {x ∈ N | x2 + 3x+ 1 = 0} is an empty set.

9


Subset. A is called a subset of B if each element in A must be in B, denoted by A ⊆ B or B ⊇ A.

A = B means A ⊆ B and B ⊇ A.

Example. Let

A = {m, a, t, h, e,m, a, t, i, c, s}, B = {m, a, t, h, e, i, c, s}.

If x ∈ B, then x ∈ A. Therefore,B ⊆ A. Also, if x ∈ A, then x ∈ B. Therefore,A ⊆ B, i.e., A = B.

Union. Union of sets A and B, denoted by A ∪B is the set of all elements which belong to A or to B, i.e.,

A ∪B = {x | x ∈ A or x ∈ B}.

Intersection. Intersection of sets A and B, denoted by A ∩ B is the set of all elements which belong to

both A and B, i.e.

A ∩B = {x | x ∈ A and x ∈ B}.

Complement. Complement of A in B, denoted by B\A is the set of all elements which belong to B and

not belong to A, i.e.

B\A = {x | x ∈ B and x �∈ A}.

1.2 Real Number

Ratio of the lengths of any two line segments. In ancient Greece, once mathematicians believed that any

quantity can be described by rational numbers. Especially they believed that for any two line segments,

there must be a third line segment such that the former two line segments are m and n (m,n are integers)

times of the length of the third line segment. Hence the ratio of the lengths of any two line segments is a

rational number m/n. The proof of many geometric theorems were based on this assumption at that time.

Meanwhile some mathematicians announced that they had proved this basic assumption. But unfortunately

all the proofs were wrong.

By Pythagorean theorem, for a right angled triangle there is a 2 + b2 = c2

Let a = b = 1, then c2 = 2.

That means the ratio of lengths c/a = γ satisfies γ2 = 2.

We now prove by algebraic method that γ can not be a rational number.

Proposition 1.2.1 No rational number satisfies the equation γ 2 = 2.


Proof: We argue by contradiction. Assume that the proposition is false, then there is a rational number

γ = m/n that satisfies (mn

)2= 2,

where m,n are integers and have no common factors other than 1. Since m 2 = 2n2, m is even. Let

m = 2k, k is an integer. Since (2k)2 = 2n2, n2 = 2k2, n is even. Therefore, 2 is a common factor of m

and n, contradicted with the assumption of no common factors other than 1. Hence, the proposition is true.

Suppose that we divide all rational numbers into two sets (Dedekind cut), say A and A ′, such that

1. Any rational number belongs only one of the two sets.

2. If x ∈ A and x′ ∈ A′, then x < x′.

There are three cases:

1. A has no largest number, but A ′ has the smallest number.

2. A has the largest number, but A′ has no smallest number.

3. A has no largest number, and A ′ has no smallest number.

Example. (case 1.)

A = {x | x < 1 and x is a rational numver.},

A′ = {x′ | x′ ≥ 1 and x is a rational numver.}.

Example. (case 3.)

A = {x | x ≤ 0 or x2 < 2, x is a rational numver.},

A′ = {x′ | x′ > 0 and x′2 > 2, x′ is a rational numver.}.

If a > 0, a ∈ A, then there exists integer n, such that n > 2a+12−a2 , or 2a+1

n < 2 − a2. Then we have

2an + 1

n2 <2a+1n < 2 − a2, or (a + 1

n )2 < 2, i.e., A has no largest number. Similarly, A ′ has no smallest

number.

In case 1 and 2, the largest of A, or the smallest of A ′ is a rational number. In case 3, we define an

irrational number. The union of all rational and irrational numbers are called real numbers. The set of real

numbers is denoted by R.

To define the order of the real numbers, let the Dedekind cuts A|A ′ and B|B′ define the real numbers

α and β. If B ⊆ A, and ∃x ∈ A s.t. x /∈ B, then we call α > β, or β < α. If A = B and A ′ = B′, then we

call α = β. It is easy to see that for any α and β, α = β, α > β, or α < β. Also, the real number defined

by Dedekind cut is unique.


Let a ∈ A, a′ ∈ A′, and b ∈ B, b′ ∈ B′, then a < α < a′ and b < β < b′. If a + b < c <

a′ + b′, ∀a, a′, b, b′, the we call α + β = c. It is not difficult to see that c is unique. Similarly, other

operations can be defined.

Real number axis. On a straight line, a point O is selected as the Origin and a line segment OE (E is on

the right of O) is selected as the unit of length. Then, the set of all real numbers and the set of all the points

on the straight line can be established a one-to-one corresponding by the following rule:

• the number x = 0 corresponds to the origin O;

• the positive (negative) number x corresponds to the point P on the right (left) of O with |x| = OPOE .

The straight line is called a Real Number Axis. Proposition 1.2.1 demonstrates that the set of rational points

cannot full fill the real number axis. The point on the axis that doesn’t correspond to a rational number

corresponds to an irrational number.

Order and operations. Using the Dedekind cut, we can also define the order (”<,>”) of two real numbers,

and the operations (”+,−,×,÷”) of real numbers.

Real number set is a field. We know that for any two real numbers x and y, the operations on x and y such

as ”addition” and ”multiplication” and their inverse ”subtraction” and ”division” are closed. That means

the results of x+ y, x− y, x− y and x/y (y �= 0) are still real numbers.

In addition, the operations have the following properties:

1. Commutative rule: a+ b = b+ a

2. Commutative rule: a · b = b · a

3. Associative rule: (a+ b) + c = a+ (b+ c)

4. Associative rule: (a · b) · c = a · (b · c)

5. Distributive rule: a · (b + c) = a · b+ a · c

Therefore, real number set is a Field. (The definition of ”Field” will be learnt in the subject ”Algebra”.

In fact, the ”real number set” is the prototype of the concept ”Field”.

Real number set is an ordered field. For any two different real numbers x and y, there is an ”Ordering”

between them. That is, one of the following relations

x < y, or y < x

must hold, i.e. y bigger than x or x bigger than y. On the real number axis, x < y means the point y is on

the right of point x.

The relationship ”Ordering” has the following properties:


1. x < y and y < z, then x < z

2. x1 < y1 and x2 < y2, then x1 + x2 < y1 + y2

3. 0 < x and y < z, then xy < xz

4. x < y, then −x > −y

5. 0 < x < y, then 1x >

1y

Symbol x ≤ y means x < y or x = y.

Density of the rationals and irrationals. Symbol R denotes the set of all real numbers.

Interval notations (a < b):

• The set {x ∈ R | a < x < b} is called an open interval denoted by (a, b).

• The set {x ∈ R | a ≤ x ≤ b} is called a closed interval, denoted by [a, b].

• [a, b) = {x ∈ R | a ≤ x < b} is called an interval but not open and not closed.

• [a, b) = {x ∈ R | a ≤ x < b}. is also called an interval but not open and not closed.

• (−∞,+∞) = R is the whole axis of real numbers, it is open and closed at the same time.

• [a,+∞) = {x ∈ R | x ≥ a} and (−∞, b] = {x ∈ R | x ≤ b} are closed intervals.

• (a,+∞) = {x ∈ R | x > a} and (−∞, b) = {x ∈ R | x < b} are open intervals.

Definition 1.2.2 A set S of real numbers is said to be Dense in R, provided that any interval I = (a, b),

contains at least one member of S.

Archimedean property. For any real numberC, there is a natural number n such that C < n. We take this

property for granted.

By this property, for any given positive number ε, we can choose a natural number n such that 1ε < n.

Hence 1n < ε.

Theorem 1.2.3 (Density theorem) The set of rational numbers and the set of irrational numbers are both

dense in R.

Proof Firstly we consider the case when 0 ≤ a < b. Choose a natural numberm such that

1

m< b− a.

Let

A =

{k | k ∈ N,

k

m< b

}.


Since A has finite members, there must be a maximum, say k0. We now show a < k0

m < b.

Since k0 ∈ A, k0

m < b. Since k0 is the maximum of A, k0+1m ≥ b. Hence

k0m

≥ b− 1

m> b− (b− a) = a.

That means the rational number k0

m satisfies a < k0

m < b.

By a similar way we can find an irrational number in (a, b). Choose a natural number m such that

1m < b−a√

2, i.e. √

2

m< b− a.

Let

B =

{k | k ∈ N, k

√2

m< b

}

and k0 be the maximum of B. The same reasoning shows that the irrational number k 0

√2

m satisfies a <

k0√2

m < b. If a < 0, 0 < b, then (0, b) ⊂ (a, b). From above proof (0, b) contains rational and irrational

numbers, so interval (a, b) does.

If a < b ≤ 0. From above proof, there are a rational number r and an irrational number x that satisfy

−b < r < −a

and

−b < x < −a

Hence

a < −r < b

and

a < −x < b

Hence we complete the proof.

It is easy to see from Theorem 1.2.3 that there is at least a rational number between any two irrational

numbers, and there is at least an irrational number between any two rational numbers.

1.3 Inequalities and Algebraic Identities

At the heart of many arguments in analysis lies the problem of estimating the sizes of various quotients,

differences and sums. In order to do so the following inequalities and identities will be used frequently.

Absolute value and basic inequalities. Absolute Value of a real number x, denoted by |x| is defined by

|x| =⎧⎨⎩ x, if x ≥ 0,

−x, if x < 0.


On the real number axis, |x| is the distance from the point x to the origin 0. |x − y| is the distance

between the point x and the point y.

By the definition we have the inequality

−|x| ≤ x ≤ |x|

and the following basic proposition.

Proposition 1.3.1 Suppose that c and d are any real numbers and d ≥ 0, then

|c| ≤ d⇔ −d ≤ c ≤ d.

Proof. ”⇒”: By the definition, |c| ≤ d means

c ≤ d and − c ≤ d.

From the second inequality it follows. Hence we obtain

−d ≤ c ≤ d.

”⇒”: Assume −d ≤ c ≤ d, i.e. −d ≤ c and c ≤ d. From −d ≤ c we get −c ≤ d. Combine −c ≤ d

and c ≤ d we obtain |c| ≤ d.

Theorem 1.3.2 (Triangle inequality) For any numbers a and b,

|a+ b| ≤ |a|+ |b|.

Proof. From inequalities

−|a| ≤ a ≤ |a|, −|b| ≤ b ≤ |b|

we have

−(|a|+ |b|) ≤ a+ b ≤ (|a|+ |b|).

Let c = a+ b, and d = |a|+ |b|, then −d ≤ c ≤ d. From Proposition 3.1.1 it follows |c| ≤ d, i.e.

|a+ b| ≤ |a|+ |b|.

From this basic form of Triangle Inequality, it is easy to get the following inequalities:

|a| − |b| ≤ |a+ b|,|a| − |b| ≤ |a− b|,

and

|x− z| ≤ |x− y|+ |y − z|.


We may see these inequalities as different forms of Triangle Inequalities.

In many cases, we need to use mathematical induction to prove theorems. To introduce the principle of

mathematical induction, we first give the following definition.

Definition 1.3.1 A set S of real numbers is said to be inductive if the number 1 ∈ S, and if x ∈ S, then

x+ 1 ∈ S.

The set of natural numbers is defined as the intersection of all inductive subsets of R, denoted by A.

Let N = {1, 2, 3, · · · }. Clearly, N is inductive. Then, A ⊆ N. On the other hand, 1 ∈ N, and 1 belongs

to every inductive set. If k ∈ N, then k belongs to every inductive set. Thus, N ⊆ A. Therefore, we have

A = N.

Principle of Mathematical Induction. For each natural number n, let S(n) be some mathematical asser-

tion. Suppose that S(1) is true. Also suppose that whenever k is a natural number such that S(k) is true,

then S(k + 1) is also true. Then S(n) is true for every natural number n.

Proof. Let A ≡ {k ∈ N | S(k) is true.}. It is enough to show that A = N. A ⊆ N by definition. The

assumption means that A is an inductive subset of N. Since A is inductive, then N ⊆ A. Then A = N, i.e.,

S(n) is true for every natural number n.

Theorem 1.3.3 (Bernoulli’s inequality) For any natural number n and real number a ≥ −1,

(1 + a)n ≥ 1 + na.

Proof. Prove by mathematical induction. For n = 1, the proposition is obviously true. Assume that the

proposition is true for n = k, i.e.,

(1 + a)k ≥ 1 + ka,

we need to prove

(1 + a)k+1 ≥ 1 + (k + 1)a.

Since

(1 + a)k+1 = (1 + a)k(1 + a)

≥ (1 + ka)(1 + a)

= 1 + (k + 1)a+ ka2

≥ 1 + (k + 1)a.

which completes the proof.


Binomial formula.

(a+ b)n =

⎛⎝ n

0

⎞⎠ an +

⎛⎝ n

1

⎞⎠ an−1b+

⎛⎝ n

2

⎞⎠ an−2b2 + · · ·+

⎛⎝ n

n− 1

⎞⎠ abn−1 +

⎛⎝ n

n

⎞⎠ bn

=

n∑k=0

⎛⎝ n

k

⎞⎠ an−kbk,

where ⎛⎝ n

k

⎞⎠ =

n!

k!(n− k)!, k! = k(k − 1) · · · 2 · 1, and 0! = 1.

Difference of powers formula.

an − bn = (a− b)

n−1∑k=0

an−1−kbk.

Geometric sum formula.n−1∑k=0

γk =1− γn

1− γ, if γ �= 1.

1.4 Function

In the real world, all things are varying. Variables are used to describe the variation in some quantity

aspects of the matters we concern. In general, two or more variables for describing a same object or

phenomenon have some relationship. For example, the volume V, temperature T and the pressure P of

certain gas satisfy V = cT/P . The concept ”function” is just for this application requirement to introduce

into mathematics. It is one of the most basic concept in Mathematical Analysis.

Definition 1.4.1. Let D and R be sets. A function from D into R, denoted by f : D → R, is a rule

that assigns to each element of D a unique element of R. The set D is called the definition domain of the

function. Functions are usually denoted by letters such as f, g, F,G, and so on. If f is a function from D

into R, then the element y ∈ R that is assigned to the element x ∈ D by f is denoted by f(x) (read ”f

of x”), and is called the value of f at x (or function value at x) or the image of x under f . The elements

x ∈ D and y ∈ R are called variables, with x the independent variable and y the dependent variable. The

set {y | y = f(x), x ∈ D} is called the range of f, or the image of D under f.

The definition domain and range of function f are denoted by D(f) and R(f) or f(D) respectively.

Throughout most of this course, the sets D and R are real number sets. In this context, a function from

D into R is said to be a real-valued function of a real variable. In general, the rule for a function will be

specified by an equation, or a set of equations.

Surjective, injective and bijective. Let f be a function from set A to set B.


• f is surjective if R(f) = B (or f(A) = B).

• f is injective if for x1 �= x2 in A, f(x1) �= f(x2) in B.

• f is bijective if f is surjective as well as injective.

Example 1: f : A→ B, f(x) = x2.

(a) If A = B = R, f is neither surjective nor injective.

(b) If A = [0,∞), B = R, f is injective.

(c) If A = R, B = [0,∞), f is surjective.

(d) If A = B = [0,∞), f is bijective.

Example 2: f : A→ B, f(x) = sinx.

(a) If A = B = R, f is neither surjective nor injective.

(b) If A = R, B = [−1, 1], f is surjective.

(c) If A = [− π2 ,

π2 ], B = R, f is injective.

(d) If A = [− π2 ,

π2 ], B = [−1, 1], f is bijective.

Inverse and composition.

• If the function f : A → B is bijective, i.e. for every point y ∈ B there is exactly one point x ∈ A

such that f(x) = y. Then, function f−1 : B → A is called the inverse of f , if for each y ∈ B,

f−1(y) = x, with f(x) = y.

Example 1: f : [0,∞) → [0,∞), f(x) = x2. The inverse f−1 : [0,∞) → [0,∞) is f−1(y) =√y.

Example 2: f : [− π2 ,

π2 ] → [−1, 1], f(x) = sinx. The inverse f−1 : [−1, 1] → [−π

2 ,π2 ] is

f−1(y) = sin−1 y. .

• Let f and g be functions and R(g) ⊆ D(f). The composition of f and g denoted by f ◦ g is defined

in D(g) by

(f ◦ g)(x) = f(g(x)).

Example: Let f(y) = sin y, g(x) = x2, then

(f ◦ g)(x) = f(g(x)) = sinx2.


1.5 Exercise

1. Use mathematical induction to prove the following identities:

(a) 12 + 22 + · · ·+ n2 =n(n+ 1)(2n+ 1)

6.

(b) 13 + 23 + · · ·+ n3 = (1 + 2 + · · ·+ n)2.

(c) 20 + 21 + · · ·+ 2n−1 = 2n − 1.

2. Find a formula forn∑

j=1

j(j + 1).

3. Prove that there is at least one irrational number between any two rational numbers.

4. Prove that there are infinitely many irrational numbers between any two rational numbers.

Hint. By mathematical induction.

5. Determine whether the following sets are dense in R and justify your answer.

(a) S ={x | x = n+

1

m, n,m, are integers

}.

(b) S ={x | x = q +

√2, q ∈ Q

}.

Hint. For any a < b, consider the interval (a −√2, b−

√2)

(c) S ={x | x =

√2q, q ∈ Q

}, Q denotes the set of all rational numbers.

6. Prove that if the set S is dense in R, then any interval I = (a, b) contains infinitely many numbers

of the set S.

7. Suppose that the number a has the property that for every natural number n, a ≤ 1/n. Prove that

a ≤ 0.

8. Suppose that the number a has the property that for every natural number n, a ≥ 1 − 1n . Prove that

a ≥ 1.

9. Prove that if a > 0, x satisfies |x− a| < a

2, then x >

a

2.

10. Solve the inequality: |x+ 2| − |x| > 1.

Hint. The zeros of |x+ 2| and |x| are: x = −2 and x = 0, then we discuss three cases: x < −2, −2 ≤ x < 0and x ≥ 0, in order to remove the absolute value signs.

11. Prove the inequality:1

2· 34· · · 2n− 1

2n<

1√2n+ 1

.

12. Show that if |a− b| ≤ 1, then |a| ≤ |b|+ 1.

13. Solve the inequality: |2x+ 3| > |x− 2|.

14. (a) Prove the Cauchy’s Inequality: for any real number a and b, ab ≤ 1

2(a2 + b2).


(b) Prove that if a and b are real numbers, and n is a natural number, then ab ≤ 1

2

(na2 +

1

nb2)

.

Hint. The inequality is equivalent to:(√

na− 1√nb)2 ≥ 0.

15. Prove that if a1, · · · , an are of same sign and ai > −1, i = 1, · · · , n, then

(1 + a1)(1 + a2) · · · (1 + an) ≥ 1 + a1 + a2 + · · ·+ an.

16. Determine whether the following functions are surjective, injective, bijective, or none of them:

(a) f : R → R, f(x) = cosx.

(b) f : (−π2,π

2) → R, f(x) = tanx.

(c) f : R → R, f(x) =

⎧⎨⎩ −1, −∞ < x < 0,

1, 0 ≤ x <∞..

(d) f : (−1, 1) → (−1, 1), f(x) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

x, −1 < x < 0,

0, 0 ≤ x < 12

2x− 1, 12 ≤ x < 1.

.

(e) f : (−1, 1) → (−1, 2), f(x) =

⎧⎨⎩ x, −1 < x < 0,

x+ 1, 0 ≤ x < 1..

17. Are the following functions bijective? If so, find their inverses.

(a) f : R → R, f(x) =

⎧⎨⎩ x, x is rational,

x+ 1, x is irrational..

(b) f : (−1, 1) → (−2, 0) ∪ [1, 5), f(x) =

⎧⎨⎩ 2x, −1 < x < 0,

4x+ 1, 0 ≤ x < 1..

(c) f : (−1, 1) → (−1, 1), f(x) =

⎧⎨⎩ x2, −1 < x < 0,

−x3, 0 ≤ x < 1..

18. Find the composition f ◦ g and g ◦ f :

f(x) =

⎧⎨⎩ −x− 1, x ≤ 0,

x, x > 0,g(x) =

⎧⎨⎩ x, x ≤ 0,

−x2, x > 0.

Chapter 2

Limit Theorey

2.1 Convergence and Limit

2.1.1 Sequence

A sequence of real number is a real-valued function f(n) : N → R, i.e., a function from the set of

natural numbers to real numbers. It is used to replace f(n) with an and denoting a sequence by {an}. The

number an associated with the index n is called the nth term of the sequence.

Note: The collection of the terms from a sequence is a set. But different terms of a sequence may be a same

number.

Countable and uncountable set. A set is called countable provided it can be counted, i.e., a one-to-one

correspondence can be established between the set and the finite set {1, 2, · · · , n} or the infinite set N.

Otherwise, the set is called uncountable. The rational number set Q is countable; but the real number set R

or its subset {x ∈ R | a < x < b} is uncountable.

The set consisted of the terms from a sequence is a countable set. An uncountable set such as {x ∈R | a < x < b} can not be arranged into a sequence which includes all the numbers of the set.

Examples 2.1.1

1. an = 1n , {an} =

{1,

1

2,1

3,1

4, · · · ,

}.

2. an = 1 +

(1

2

)n

, {an} =

{1 +

1

2, 1 +

1

4, 1 +

1

8, 1 +

1

16, · · · ,

}.

3. a1 = 1, an+1 =

⎧⎨⎩ an + 1

n , if a2n ≤ 2,

an − 1n , if a2n > 2,

{an} =

{1, 2,

3

2,7

6, · · · ,

}.

4. an = 1 + (−1)n, {an} = {0, 2, 0, 2, · · · , }.

21


5. an =

(1 +

1

n

)n

, {an} =

{2,

9

4,64

27,625

256, · · · ,

}.

2.1.2 Convergence and Limit

Observing the example 1, we find that when n becomes larger and larger, the number a n = 1n becomes

closer and closer to zero. The sequence {an} has infinitely many terms and an never equals zero. But an

can be arbitrarily close to zero while the index n is sufficiently large. Similarly, the sequence in example 2

the terms an = 1+(12

)nalthough never equal 1, but an can be arbitrarily close to 1 while n is sufficiently

large.

Definition 2.1.1 A sequence {an} is said to be convergent and converge to the number a if for any positive

number ε (or say arbitrarily small number ε > 0) there exists a natural numberN (∃N ) such that

|an − a| < ε ∀n ≥ N.

The number a is called the limit of the sequence {an}, where ”∀” means ”for all”, or we write as

Given ε > 0, ∃N > 0, s.t. |an − a| < ε ∀n ≥ N.

Notation: A sequence {an} converges to number a is denote by

limn→∞ an = a, or an → a.

Geometrical explanation. The inequality

|an − a| < ε ∀n ≥ N.

means that all the terms an with index n ≥ N are inside the interval (a− ε, a+ ε), equivalently, there are

only a finite number of terms an outside the interval.

Proposition 2.1.2 A sequence {an} converges to a number a if and only if for any ε > 0, only a finite

number of terms of an are outside the interval (a− ε, a+ ε).

A sequence is said to be divergent if the sequence does not converge. The sequence in example 4 is diver-

gent.

Definition 2.1.3 A sequence {xn} is said to tend to +∞ (or −∞) if for any positive numberM (or negative

number −M ) there exists a natural numberN such that

xn > M (orxn < −M), ∀n ≥ N

Chapter 2: Limit Theory 23

and denoted by

limn→∞xn = +∞, (or lim

n→∞xn = −∞).

If there exists a natural number N such that

|xn| > M, ∀n ≥ N,

then we say that

limn→∞ xn = ∞.

Examples 2.1.2 Prove limn→∞

1

n= 0.

Proof. According to the definition, for any given ε > 0 we need to find N such that∣∣∣∣ 1n − 0

∣∣∣∣ < ε ∀n ≥ N.

⇔ 1

n< ε ∀n ≥ N.

⇔ 1

ε< n ∀n ≥ N.

From this, we see that when we chooseN =[1ε

]+ 1 ([x] means the integer part of x.), then

∣∣∣∣ 1n − 0

∣∣∣∣ < ε ∀n ≥ N.

Thus we complete the proof.

Note: The numberN that we are going to find is a function of ε. It is not unique. In the example,N =[1ε

]+k

is a right choice with any positive integer k.

Examples 2.1.3 Prove limn→∞

[1 +

(1

2

)n]= 1.

Proof. For any ε > 0, we need to find N such that∣∣∣∣[1 +

(1

2

)n]− 1

∣∣∣∣ < ε, ∀n ≥ N.

⇔(1

2

)n

< ε ∀n ≥ N.

⇔ n ln

(1

2

)< ln ε ∀n ≥ N.

⇔ −n ln(1

2

)> − ln ε ∀n ≥ N.

⇔ n >− ln ε

− ln(12

) = ln ε

ln(12

) , ∀n ≥ N.


From this, we see that when we chooseN =

[ln ε

ln( 12 )

]+ 1 then

∣∣∣∣[1 +

(1

2

)n]− 1

∣∣∣∣ < ε, ∀n ≥ N.

Examples 2.1.4 Let sequence {an} be defined as

a1 = 1, an+1 =

⎧⎨⎩ an + 1

n , if a2n ≤ 2,

an − 1n , if a2n > 2,

Prove limn→∞ an =

√2.

Proof. First we prove by mathematical induction the inequality

∣∣∣an −√2∣∣∣ ≤ 1

n− 1, n ≥ 2.

When n = 2, the inequality is true due to

∣∣∣a2 −√2∣∣∣ = ∣∣∣2−√

2∣∣∣ ≤ 1.

Assume that the inequality is true for n = k, i.e.,

∣∣∣ak −√2∣∣∣ ≤ 1

k − 1. (2.1.1)

We need to show that the inequality is also true for n = k + 1. From (2.1.1) we get

− 1

k − 1< ak −

√2 <

1

k − 1.

There are two cases:

(i) − 1

k − 1< ak −

√2 ≤ 0, (2.1.2)

(ii) 0 < ak −√2 <

1

k − 1. (2.1.3)

In case (i), by the definition of the sequence,

ak+1 = ak +1

k.

Hence

ak+1 −√2 = ak −

√2 +

1

k.

Adding 1k to (2.1.2) we get

1

k− 1

k − 1< ak+1 −

√2 ≤ 1

k,

or,

− 1

k(k − 1)< ak+1 −

√2 ≤ 1

k.


Since k ≥ 2, we have

− 1

k< ak+1 −

√2 ≤ 1

k,

i.e., ∣∣∣ak+1 −√2∣∣∣ ≤ 1

k.

In case (ii), by the definition of the sequence,

ak+1 = ak − 1

k.

Hence

ak+1 −√2 = ak −

√2− 1

k.

Subtracting 1k from (2.1.3) we get

− 1

k< ak+1 −

√2 <

1

k − 1− 1

k,

or,

− 1

k< ak+1 −

√2 <

1

k(k − 1).

Since k ≥ 2, we have

− 1

k< ak+1 −

√2 <

1

k,

i.e., ∣∣∣ak+1 −√2∣∣∣ ≤ 1

k.

We have proved the inequality ∣∣∣an −√2∣∣∣ ≤ 1

n− 1.

Thus for any ε > 0, we choose N =[1ε

]+ 2, then∣∣∣an −√

2∣∣∣ ≤ ε, ∀n ≥ N.

Examples 2.1.5 Prove that the sequence {0, 2, 0, 2, · · · } is divergent.

Proof. We argue by contradiction. Assume that limn→∞ an = a. According to the definition of conver-

gence, we choose ε = 12 , then there exists a natural numberN such that

|an − a| < 1

2∀n ≥ N.

Therefore, the neighboring two terms aN and aN+1 satisfy

|aN − aN+1| ≤ |aN − a|+ |aN+1 − a| < 1

2+

1

2= 1.

But, according to the definition of the sequence

|aN − aN+1| = 2.

The contradiction shows that the sequence can’t converge.


2.1.3 Uniqueness of Limit

Proposition 2.1.4 A sequence can not converge to two limits.

Proof. We argue by contradiction. Let limn→∞ an = a and also lim

n→∞ an = a′, a′ �= a. Take ε = 12 |a − a′|.

According to the definition of convergence,

an → a ⇒ ∃N1 s.t. |an − a| < ε, ∀n ≥ N1,

an → a′ ⇒ ∃N2 s.t. |an − a′| < ε, ∀n ≥ N2.

Define N = max{N1, N2}, then we have

|an − a| < ε and |an − a′| < ε, ∀n ≥ N.

Thus,

|a− a′| ≤ |an − a|+ |an − a′| < 2ε = |a− a′| < ε,

which is impossible, hence a sequence can’t converge to two limits.

2.1.4 Boundedness of Convergent Sequence

Bounded above. A sequence {an} is said to be bounded above if for all an there is a number B such that

an ≤ B.

The number B is called an upper bound of the sequence. Obviously, if B is an upper bound of a sequence,

then any numberB ′ > B is also an upper bound of the sequence.

Bounded below. A sequence {an} is said to be bounded below, if for all an there is a numberA such that

A ≤ an.

The number A is called a lower bound of the sequence. If A is a lower bound of a sequence, then any

numberA′ < A is also a lower bound of the sequence.

Bounded sequence. A sequence {an} is said to be bounded if it is bounded above as well as bounded

below. Therefore there are numbers A and B such that for all an

A ≤ an ≤ B.

Define M = max{|A|, |B|}, then

|an| ≤M,


or

−M ≤ an ≤M.

Lemma 2.1.5 Every convergent sequence is bounded.

Proof. Let {an} be a convergent sequence and converges to a. Taking ε = 1, it follows from the definition

of convergence that there exists a natural numberN such that

|an − a| < 1, ∀n ≥ N.

Then we have

|an| − |a| ≤ |an − a| < 1, ∀n ≥ N,

or

|an| < 1 + |a|, ∀n ≥ N.

Let

M = max{1 + |a|, |a1|, |a2|, · · · , |aN−1|},

Then

|an| < M, ∀n ∈ N.

Lemma 2.1.6 Suppose that

limn→∞ bn = b �= 0.

Then there is a natural number N such that

|bn| > |b|2, ∀n ≥ N.

Proof. Take ε = |b|2 . Then there exists a natural numberN such that

|bn − b| < |b|2, ∀n ≥ N,

then we have

|b| − |bn| ≤ |bn − b| < |b|2, ∀n ≥ N,

or

|bn| > |b|2, ∀n ≥ N.


2.1.5 Operations on Convergent Sequences

For some simple sequences, we can use the ”ε−N” definition to obtain their limits. For example, it is

easy to obtain the limits of the following sequences

an =

(3

4

)n

, bn =1√n

by ”ε−N” definition, but it is difficult to obtain the limit of the sequence

cn =1

1 +(34

)n +

(1 +

1√n

)2

by ”ε−N” definition.

Theorem 2.1.7 Suppose that

limn→∞ an = a, and lim

n→∞ bn = b.

Then

(i) limn→∞(an + bn) = lim

n→∞ an + limn→∞ bn = a+ b.

(ii) limn→∞(an · bn) = lim

n→∞ an · limn→∞ bn = a · b.

(iii) If bn �= 0, ∀n, and b �= 0, then, limn→∞

anbn

=limn→∞ an

limn→∞ bn

=a

b.

Proof of (i). Let ε > 0. We need to find a natural numberN such that

|(an + bn)− (a+ b)| < ε, ∀n ≥ N.

Since an → a, we can choose a natural numberN1 such that

|an − a| < ε

2, ∀n ≥ N1.

Since bn → b, we can choose a natural numberN2 such that

|bn − b| < ε

2, ∀n ≥ N2.

Let N = max{N1, N2}, then

|(an + bn)− (a+ b)| ≤ |an − a|+ |bn − b| < ε

2+ε

2= ε, ∀n ≥ N.


Proof of (ii). Let ε > 0. We need to find a natural numberN such that

|anbn − ab| < ε, ∀n ≥ N.


Since

anbn − ab = anbn − anb+ anb− ab

= an(bn − b) + (an − a)b,

then

|anbn − ab| < |an||bn − b|+ |b||an − a|.

Since {an} converges, by Lemma 2.1.5 there is a numberM such that

|an| < M, ∀n.

Hence,

|anbn − ab| < M |bn − b|+ |b||an − a|.

Since bn → b, we can choose a natural numberN1 such that

|bn − b| < ε

2M∀n ≥ N1.

If b = 0, then

|anbn − ab| < M |bn − b| < ε

2< ε ∀n ≥ N1,

which gives the proof. If b �= 0, since an → a, we can choose a natural numberN2 such that

|an − a| < ε

2|b| ∀n ≥ N2.

Let N = max{N1, N2}, then

|anbn − ab| < Mε

2M+ |b| ε

2|b| = ε ∀n ≥ N.

We complete the proof.

Proof of (iii). Using (ii), it suffices to prove that

limn→∞

1

bn=

1

b.

Since1

bn− 1

b=b− bnbbn

, .

according to Lemma 2.1.6, we can chooseN1 such that

|bn| > |b|2, ∀n ≥ N1.

Thus, ∣∣∣∣ 1bn − 1

b

∣∣∣∣ =∣∣∣∣b− bnbbn

∣∣∣∣ < 2

|b|2 |bn − b| ∀n ≥ N1.


Since bn → b, we can chooseN2 such that

|bn − b| < |b|22ε ∀n ≥ N2.

Let N = max{N1, N2}, then∣∣∣∣ 1bn − 1

b

∣∣∣∣ < 2

|b|2 |bn − b| < 2

|b|2|b|22ε = ε ∀n ≥ N,

which completes the proof.

Corollary 2.1.8

limn→∞(αan + βbn) = α lim

n→∞ an + β limn→∞ bn,

limn→∞

m∑k=0

ckxkn =

m∑k=0

ck

(limn→∞xn

)k.

Now it is easy to obtain the limit of the sequence

cn =1

1 +(34

)n +

(1 +

1√n

)2

mentioned above:

limn→∞ cn =

1

limn→∞[1 +(34

)n] +(1 + lim

n→∞1√n

)2

= 1 + 1 = 2.

2.1.6 Squeezing Principle

Lemma 2.1.9 Suppose that dn ≥ 0 for all n ≥ N0, where N0 is some natural number and limn→∞ dn = d.

Then d ≥ 0.

Proof. If d < 0, since dn → d, for ε = |d|/2 there is a numberN such that

d− |d|2< dn < d+

|d|2

=d

2< 0 ∀n ≥ N.

which contradicts with the assumption dn ≥ 0 for all n ≥ N0.

Theorem 2.1.10 Suppose an → a, bn → b, cn → c and

an ≤ cn ≤ bn ∀n ≥ N0,

where N0 is some natural number. Then

a ≤ c ≤ b.

Proof. Let dn = cn − an. Since dn ≥ 0 for all n ≥ N0, by Lemma 2.1.9 dn → d ≥ 0. Hence

limn→∞(cn − an) = lim

n→∞ cn − limn→∞ an = c− a = d ≥ 0,


i.e., c ≥ a. Similarly we prove, c ≤ b.

Corollary 2.1.11 Suppose that a ≤ cn ≤ b for all n ≥ N0, where N0 is some natural number and cn → c.

Then a ≤ c ≤ b.

Theorem 2.1.12 Suppose that sequences {an}, {bn} and {cn} satisfy

an ≤ cn ≤ bn ∀n ≥ N0,

where N0 is some natural number, and

limn→∞ an = lim

n→∞ bn = l,

then

limn→∞ cn = l.

Proof. By the assumptions an → l and bn → l, for any given ε > 0 there exist N1 and N2 such that

|bn − l| < ε ∀n ≥ N1,

|an − l| < ε ∀n ≥ N2.

It follows than

bn < l + ε ∀n ≥ N1,

an > l − ε ∀n ≥ N2.

Let N = max{N0, N1, N2}, from the assumption an ≤ cn ≤ bn we get

l − ε < an ≤ cn ≤ bn < l + ε ∀n ≥ N.

That is

|cn − l| < ε ∀n ≥ N,

or,

limn→∞ cn = l.

Note: Here we do not need to assume that {cn} converges.

The Squeezing Principle will be used frequently onwards. It is very useful to find the limits of a

sequence with complicated expression.

Examples 2.1.6 Show that

limn→∞[(n+ 1)k − nk] = 0, 0 < k < 1.


Proof.

0 < (n+ 1)k − nk = nk[(1 +

1

n

)k − 1]< nk

[(1 +

1

n

)− 1]=

1

n1−k.

Since limn→∞

1n1−k = 0, by Squeezing Principle lim

n→∞[(n+ 1)k − nk] = 0.

Examples 2.1.7 Let

cn =1√

n2 + 1+

1√n2 + 2

+ · · ·+ 1√n2 + n

,

show that limn→∞cn = 1.

Proof.

an ≡ n√n2 + n

< cn <n√

n2 + 1≡ bn.

Since limn→∞an = lim

n→∞bn = 1, by Squeezing Principle limn→∞cn = 1.

2.2 Some Important Theorems in Limit Theory

We have mentioned some properties of the real number set in Chapter 1 such as the following: The real

number set is an ordered field; the real number set is dense in itself, namely, there is another real number

in between any two real numbers. But these properties are also shared by the rational number set. In this

section we will introduce the property of completeness (or continuity) of the real number set, which is not

owned by the rational number set.

The property of completeness can be expressed in any one of the following seven equivalent theorems

that we will mention. Among these theorems we can take any one of them as an Axiom and then the

other six theorems can be proved accordingly. These seven theorems compose the theoretical basis of

mathematical analysis.

2.2.1 Dedekind Theorem

Theorem 2.2.1 (Dedekind Theorem) Suppose that A and A ′ is a cut for the real number set (Dedekind

cut), then this cut generates a real number, which is the largest number of A, or the smallest number of A ′.

Proof. Let B and B ′ be the sets containing all rational numbers of A and A ′ respectively. Then B and B ′

is a cut for the rational number set, which defines a real number a: a ∈ A or a ∈ A ′. If a ∈ A, then a is the

largest number of A, since otherwise, there exists a real number b > a, then we can find a rational number

r such that a < r < b, which contradicts with the conditions of the cut B and B ′. Similarly, if a /∈ A, then

a ∈ A′ and a is the smallest number of A′.

This theorem shows that the real number axis is complete, i.e., the real number axis has no holes. If

we just admit this property of the real number set without proof, then the above theorem is usually called

Dedekind Continuous Axiom.


2.2.2 Least Upper Bound Theorem

It is known that there must be a maximum and minimum within a nonempty set of finite numbers.

But a set of infinitely many numbers may not have maximum and minimum. For examples, the set A ={r | r = 1

n , n ∈ N}

has no minimum; the set B ={r | r = 2− 1

n , n ∈ N}

has no maximum; the set

C = A ∪B has neither maximum nor minimum.

Definition 2.2.2 A nonempty set S of real numbers is said to be bounded above provided that there is a

number C having the property

x ≤ C, ∀x ∈ S.

Such a number C is called an upper bound of S.

Corollary 2.2.3 If number C is an upper bound of set S, then any number C ′ ≥ C is also an upper bound

of S.

Consider the example A ={r | r = 1

n , n ∈ N}

. The number 1 is an upper bound and also the maxi-

mum of set A. Because 1 ∈ A, any upper bound C of A satisfies C ≥ 1. Hence 1 is the minimum of the

set of all upper bounds, or, 1 is the least upper bound of A.

Now consider the example B ={r | r = 2− 1

n , n ∈ N}

. The set has no maximum, but bounded

above. The number 2 is an upper bound of set B. For any number C < 2, because 2− C > 0, we can find

a natural number n such that 1n < 2 − C. Hence C < 2 − 1

n and then C is not an upper bound of set B.

That means the number 2 is the least upper bound of set B.

Both the examples give an affirmative answer to the question. In fact, we have the following Axiom for

general set bounded above.

Theorem 2.2.4 (Least Upper Bound Theorem) Suppose that S is a nonempty set of real numbers that is

bounded above. Then among the set of upper bound for S there is a smallest, or least upper bound.

Proof. Let A′ be the set of all upper bounds of S, and A be the set of all other numbers. Then A and A ′

is a cut for the real numbers. By Dedekind Theorem, it generates a real number b. b is an upper bound of

S, b ∈ A′, and b is the smallest number of A′.

Similarly, if we just admit this property of the real numbers without proof, then this theorem can also

be used as the basis of the real number system, and the theorem can be called the Completeness Axiom.

Using this theorem, we can derive the Dedekind Theorem, i.e., the two theorems are equivalent.

Proof of Dedekind Theorem. Suppose that A and A′ is a cut for the real numbers. If A has the largest

number, then we have the proof. If A has no largest number, then by Least Upper Bound Theorem, A ′ has

the smallest number.


If b is the least upper bound of set S, we denote it by

b = l.u.b. S

Sometimes the least upper bound of S is also called the supremum of S and is denoted by

b = supS

According to the definition of least upper bound, it is obvious that

(i) b is the least upper bound of set S if and only if b is an upper bound and for any upper bound d of S,

d ≥ b.

(ii) b is the least upper bound of set S if and only if b is an upper bound of S and ∀ε > 0, ∃x 0 ∈ S such

that x0 > b− ε.

Definition 2.2.5 A nonempty set S of real numbers is said to be bounded below provided that there is a

number C having the property that

x ≥ C, ∀x ∈ S.

Such a number C is called a lower bound of S.

From the definition, if C is a lower bound of S and C ′ < C, then C ′ is also a lower bound of S. By the

Completeness Axiom for the set bounded above, we have the following corollary.

Corollary 2.2.6 Suppose that S is a nonempty set and bounded below. Then among the set of lower bounds

for S there is a greatest lower bound.

We denote the greatest lower bound b of S by

b = g.l.b. S

Sometimes we call the greatest lower bound of S the infimum of S and denote it by

b = inf S

According to the definition of greatest lower bound, it is obvious that

(i) b is the greatest lower bound if and only if b is a lower bound and for any lower bound d of S, b ≥ d.

(ii) b is the greatest lower bound if and only if b is a lower bound and ∀ε > 0, ∃x 0 ∈ S such that x0 < b+ε.

In summary, the completeness property of real numbers can also be expressed in the following theorem.

Theorem 2.2.7 (Supremum and Infimum Theorem). A nonempty set of real numbers bounded above has

the supremum; A nonempty set of real numbers bounded below has the infimum.


Rational number set is not complete: S is a subset of Q and is bounded above, but S may not have

supremum in Q.

Examples 2.2.1 Consider the set S = {r ∈ Q | r ≤ √2}. S is bounded above, but it has no supremum

(least upper bound).

Proof. We argue by contradiction. Assume that S has a least upper bound b = mn . We have proved that

x2 = 2 has no solution in Q, hence mn �= 2. Since b is an upper bound of S, b >

√2. By the density

theorem, there is a rational number b ′ in interval (√2, b), i.e.

√2 < b′ < b,

which means b′ also an upper bound of S and b ′ < b. Hence b is not the least upper bound. This contradic-

tion shows no least upper bound of S in Q.

2.2.3 Monotone Convergence Theorem

Definition 2.2.8 A sequence {an} is said to be monotonically increasing provided

an+1 ≥ an, ∀n ∈ N,

{an} is said to be monotonically decreasing provided

an+1 ≤ an, ∀n ∈ N,

Both cases are called monotone.

Theorem 2.2.9 A sequence {an} that is monotonically increasing and bounded above or is monotonically

decreasing and bounded below must converge.

Proof. Let’s first assume that {an} is monotonically increasing and bounded above. By the Completeness

Axiom, the sequence {an} has a least upper bound a = l.u.b. {an}. We claim that an → a. It means that

for any given ε > 0, we need to find a natural numberN such that

a− ε < an < a+ ε, ∀n ≥ N.

Since a is the least upper bound of {an}, by the definition of least upper bound, there is a term of {an}, say

aN , such that

a− ε < aN .

The indexN of this term aN is just what we want to find. Since an is monotonically increasing,

an ≥ aN , ∀n ≥ N.


On the other hand, a is an upper bound. Hence we obtain

a− ε < aN ≤ an < a+ ε, ∀n ≥ N,

which completes the proof of

limn→∞ an = a = l.u.b. {an},

Similarly if {an} is monotonically decreasing and bounded below, then we have

limn→∞ an = a = g.l.b. {an}.

In section 1, we have proved that a convergent sequence must be bounded. Therefore, a theorem

connects the three concepts ”bounded” ”monotone” and ”convergent” can be expressed as follows.

Theorem 2.2.10 (Monotone Convergence Theorem) A monotone sequence converges if and only if it is

bounded.

Examples 2.2.2 Show that the sequence

sn =

n∑k=1

1

2kk, n = 1, 2, · · · ,

converges.

Proof. Clearly, sn is monotonically increasing, and sn ≤n∑

k=1

1

2k≤ 1, i.e., sn is bounded above. By

Monotone Convergence Theorem, sn converges.


sn =

n∑k=1

1

k, n = 1, 2, · · · ,

diverges.

Proof. Clearly, sn is monotonically increasing. Also, we have

s2 = 1 +1

2≥ 1 +

1

2,

s4 = s2 +1

3+

1

4≥ s2 +

1

4+

1

4= 1 +

2

2,

s8 = s4 +1

5+ · · ·+ 1

8≥ s4 +

4

8≥ 1 +

3

2.

If s2n−1 ≥ 1 +n− 1

2, then

s2n = s2n−1 +1

2n−1 + 1+ · · ·+ 1

2n

≥ s2n−1 +1

2≥ 1 +

n− 1

2+

1

2= 1 +

n

2.


By Mathematical Induction, it is true for any n, or

sn ≥ 1 +1

2log2 n, ∀n.

Since limn→∞ log2 n = +∞, {sn} diverges.


an =

(1 +

1

n

)n

, n = 1, 2, · · · ,

converges.

Proof. According to the Monotone Convergence Theorem, we only need to prove that the sequence is

bounded and monotone. Expand an:

an = 1 + n · 1n+n(n− 1)

2 · 11

n2+ · · ·+ n!

k!(n− k)!

1

nk+ · · ·+ n!

n!(n− n)!

1

nn

= 1 + 1 +1

2!

(1− 1

n

)+

1

3!

(1− 1

n

)(1− 2

n

)+ · · ·+

1

n!

(1− 1

n

)· · ·(1− n− 1

n

)

Similarly, the expansion of an+1 is:

an+1 =

(1 +

1

n+ 1

)n+1

= 1 + 1 +1

2!

(1− 1

n+ 1

)+

1

3!

(1− 1

n+ 1

)(1− 2

n+ 1

)+ · · ·+

1

(n+ 1)!

(1− 1

n+ 1

)· · ·(1− n

n+ 1

)

Comparing term by term, we get

an < an+1.

Hence the sequence is monotonically increasing. According to the above expansion of a n, we see that

2 < an < 1 + 1 +1

2+

1

22+ · · ·+ 1

2n−1< 3,

which means that {an} is bounded. Therefore the Monotone Convergence Theorem implies that the se-

quence {an} converges. The limit of this sequence is one of the important constants of mathematics and is

denoted by

e = limn→∞

(1 +

1

n

)n

.

Examples 2.2.5 Let a1 = 1, and

an+1 =1 + an2 + an

, n = 1, 2, · · · ,


Show that {an} converges and find the limit.

Proof. It is easy to see that {an} > 0, ∀n, and a2 < a1. Since

an+2 − an+1 =1 + an+1

2 + an+1− 1 + an

2 + an

=an+1 − an

(2 + an)(2 + an+1)< 0, ∀n,

{an} is monotonically decreasing and bounded below. {an} converges by the Monotone Convergence

Theorem. Let limn→∞ an = a, then

a =1 + a

2 + a,

and a =−1 +

√5

2.

2.2.4 Nested Interval Theorem

Nested sequence of intervals. A sequence of intervals {In} that satisfies

In+1 ⊆ In

for every natural number n is called nested sequence of intervals.

Closed Sets. A subset S of R is said to be closed provided that whenever {an} is a sequence in S that

converges to a number a, then a ∈ S.

Open Sets. A subset S of R is said to be open provided that for any x ∈ S, there exists a δ > 0 such that

(x − δ, x + δ) ⊆ S. (x − δ, x + δ) is called a neighborhood of x, and a such point x is called an interior

point.

Theorem 2.2.11 (Nested Interval Theorem) Suppose that a sequence of intervals {In} satisfies

(i) All intervals In = [anbn] are closed intervals.

(ii) The sequence of interval {In} is nested, i.e. an ≤ an+1 < bn+1 ≤ bn for every n.

(iii) limn→∞(bn − an) = 0.

Then there is exactly one point C that belongs to all intervals In, and both the sequences {an} and {bn}converge to C.

Proof. From the assumption (ii), the sequences {an} and {bn} are both monotone and bounded. By the

Monotone Convergence Theorem,

limn→∞ an = l.u.b. {an},


and

limn→∞ bn = g.l.b. {bn}.

From the assumption (iii),

limn→∞(bn − an) = lim

n→∞ bn − limn→∞ an = 0.

Denote C = limn→∞ bn = lim

n→∞ an. Then

C = l.u.b. {an} = g.l.b. {bn}.

Namely C is an upper bound of an as well as a lower bound of bn. Hence

an ≤ C ≤ bn ∀n.

Since all In are closed intervals, the above inequality means C ∈ In, ∀n.

Note: If the assumption (i) of the theorem is changed from closed intervals to open intervals. The conclu-

sion may not be true. For example, for In =(0, 1

n

)the nested sequence of intervals {In} satisfies the

assumptions (ii) and (iii). But there doesn’t exist any point that belongs to all In, since

limn→∞ an = lim

n→∞ bn = 0,

and 0 �∈ In, ∀n.

2.2.5 Heine-Borel Theorem

Theorem 2.2.12 (Heine-Borel Theorem) Let Iλ, λ ∈ Λ ⊆ R, be open intervals and the collection of sets

E = {Iλ | λ ∈ Λ} be an open cover of [a, b], i.e., ∀x ∈ [a, b], ∃λ ∈ Λ s.t. x ∈ Iλ. Then E has a finite

subcover, i.e., ∃E1 ⊆ E, E1 is an open cover of [a, b] and the number of sets in E1 is finite.

Proof. Suppose that [a, b] can not be covered by a finite number of open intervals. Divide [a, b] into [a, a+b2 ]

and [ a+b2 , b], then one of these two intervals can not be covered by a finite number of open intervals. Denote

this interval by [a1, b1] and divide [a1, b1] into two subintervals [a1,a1+b1

2 ] and [ a1+b12 , b1]. Then one of

these two intervals can not be covered by a finite number of open intervals. Continue in this way, we get a

series of intervals [a1, b1], [a2, b2], [a3, b3], · · · . For any n, [an, bn] can not be covered by a finite number

of open intervals, and the sequence an and bn satisfy

an ≤ an+1 ≤ bn+1 ≤ bn,

limn→∞(bn − an) = lim

n→∞1

2(bn−1 − an−1) = · · · = lim

n→∞1

2n(b − a) = 0.

From Nested Interval Theorem, ∃c = limn→∞ an = lim

n→∞ bn. Thus, ∃λ s.t. c ∈ Iλ. Since Iλ is open, ∃ε s.t.

(c− ε, c+ ε) ⊆ Iλ.


Since limn→∞ an = lim

n→∞ bn = c, for this ε, ∃N s.t.

|an − c| < ε

2, |bn − c| < ε

2, ∀n ≥ N,

or,

[an, bn] ⊆ (c− ε, c+ ε), ∀n ≥ N,

i.e., [an, bn] can be covered by one open interval Iλ, which is a contradiction.

This theorem is also called Finite Cover Theorem.

Examples 2.2.6 Let In = (1

n, 2), and E = ∪∞

n=1In. Then, E is an open cover of (0, 1], but E has no finite

cover.

2.2.6 Bolzano-Weierstrass Theorem

Subsequence. Consider a sequence {an}. Let {nk} be a sequence of natural number and satisfies

n1 < n2 < n3 < · · ·

Then the sequence {ank} is called a subsequence of {an}.

Proposition 2.2.13 Suppose that the sequence {an} converges to number a, then any subsequence of {an}also converges to a.

Theorem 2.2.14 (Bolzano-Weierstrass Theorem) Every sequence in a bounded and closed interval [a, b]

has a subsequence that converges to a point in [a, b].

Proof. Let {xn} ⊆ [a, b]. We claim that there is a point c ∈ [a, b] s.t. there are infinite many terms of

{xn} in any neighborhood of c. Suppose that this is not true, then ∀x ∈ [a, b], ∃δ x s.t. there are only a

finite number of terms of {xn} in (x− δx, x+ δx) ≡ Ix. Then E = {Ix | x ∈ [a, b]} forms an open cover

of [a, b]. From the Finite Cover Theorem, there exists a finite subcover E 1. Since there are only a finite

number of terms in each Ix ⊆ E1, the total terms in E1 is also finite, i.e., only a finite number of terms in

[a, b], which is a contradiction. Now, ∀k, take one term of {xn} from (c − 1/k, c+ 1/k) and denote it by

xnk. Then

|c− xnk| < 1

k, k = 1, 2, · · · ,

and

limk→∞

xnk= c.


2.2.7 Cauchy Convergence Criterion

Cauchy sequence. Sequence {an} is said to be a Cauchy sequence provided that for each positive number

ε, there is a natural numberN such that

|an − am| < ε, ∀n,m ≥ N.

Examples 2.2.7 Sequence { 1n} is a Cauchy sequence since∣∣∣∣ 1n − 1

m

∣∣∣∣ < 1

n, ∀n > m,

hence for any ε > 0, choose N =[1ε

]+ 1 then 1

n < ε if n ≥ N and∣∣∣∣ 1n − 1

m

∣∣∣∣ < ε, ∀n,m ≥ N.

Examples 2.2.8 Sequence {an} = {(−1)n} is not a Cauchy sequence.

Proof. Let ε0 = 1, then for anyN , let n0 = N ≥ N and m0 = N + 1 ≥ N , but

|an0 − am0 | = |(−1)N − (−1)N+1| = 2 > ε0,

i.e., {an} is not a Cauchy sequence.

Theorem 2.2.15 (Cauchy Convergence Criterion Theorem) A sequence converges if and only if it is a

Cauchy sequence.

Proof. Necessity part: i.e., assume that {an} is a convergent sequence, we want to show that {an} is a

Cauchy sequence. Given ε > 0, we need to findN such that

|an − am| < ε, ∀n,m ≥ N.

Assume an → a, we may choose a natural numberN s.t.

|ak − a| < ε

2, ∀k ≥ N.

Thus, if n,m ≥ N , the Triangle Inequality implies

|an − am| = |(an − a) + (a− am)|≤ |an − a|+ |a− am| < ε

2+ε

2= ε.

which means that {an} is a Cauchy sequence.

Sufficiency part: i.e., assume that {an} is Cauchy sequence, we want to show that {an} is a convergent

sequence. We first show that Cauchy sequence is bounded. Take ε = 1, then there is a N such that

|an − am| < 1, ∀n,m ≥ N.


In particular

|an − aN | < 1, ∀n ≥ N.

Consequently, we see that

|an| ≤ |aN |+ 1, ∀n ≥ N.

Then for all n,

|an| ≤ max{|aN |+ 1, |a1|, |a2|, · · · , |aN−1|},

i.e., {an} is a bounded sequence. Hence, by Bolzano-Weierstrass Theorem {an} has a subsequence {ank}

that converges to a. We now prove that the whole sequence {an} converges to a. Since {an} is a Cauchy

sequence, we can choose a N such that for any given ε > 0

|an − am| < ε

2, ∀n,m ≥ N.

Since {ank} → a, there is a natural numberK such that

|ank− a| < ε

2, ∀k ≥ K.

ChooseK such that when k ≥ K,nk ≥ N . Then it follows that if n ≥ N ,

|an − a| = |(an − ank) + (ank

− a)|≤ |an − ank

|+ |ank− a| < ε

2+ε

2= ε.

which means that {an} converges to a.

Examples 2.2.9 Let

an =n∑

k=1

sin k

2k,

show that {an} converges.

Proof. Let m > n, then

|am − an| =∣∣∣ sin(n+ 1)

2n+1+ · · ·+ sinm

2m

∣∣∣≤ 1

2n+1+ · · ·+ 1

2m<

1

2n.

For given ε > 0, let N = [log2 1/ε] + 1, then

|am − an| < ε, ∀m,n ≥ N,

i.e., {an} is a Cauchy sequence and converges.

Example 2.2.10 The sequence

Sn = 1 +1

2+

1

3+ · · ·+ 1

n, n = 1, 2, · · ·


diverges.

Proof. For m > n

|Sn − Sm| = 1

n+ 1+

1

n+ 2+ · · ·+ 1

m.

Let ε0 = 13 , then for anyN , let n0 = N ≥ N andm0 = 2N ≥ N , but

|Sm0 − Sn0 | =1

N + 1+

1

N + 2+ · · ·+ 1

2N> N · 1

2N=

1

2> ε0,

i.e., {Sn} is not a Cauchy sequence, by Cauchy Convergence Criterion, it diverges.

So far we have introduced seven important theorems:

1. Dedekind Theorem

2. Least Upper Bound Theorem

3. Monotone Convergence Theorem

4. Nested Interval Theorem

5. Heine-Borel (Finite Cover) Theorem

6. Bolzano-Weierstrass Theorem

7. Cauchy Convergence Criterion Theorem

These seven theorems are all equivalent. To show this, we only need to prove Dedekind Theorem or Least

Upper Bound Theorem by using Cauchy Convergence Criterion Theorem, which is given below.

Proof of Least Upper Bound Theorem from Cauchy Convergence Criterion Theorem. Let the real

number set {xn} be nonempty and bounded above. We need to prove that {x n} has the least upper bound.

If {xn} has only finite elements or has maximum, then obviously it has the least upper bound. Hence, here

we assume that {xn} has infinite elements and no maximum.

Let b be an upper bound of {xn}. Take x1 to be a0 and b to be b0. Subdivide the interval [a0, b0] into

two subinterval [a0, a0+b02 ] and [ a0+b0

2 , b0]. Because {xn} has no maximum, the interval [ a0+b02 , b0] can’t

have finite but not zero elements of the set {xn}. That means there are only two possibilities: [ a0+b02 , b0]

has none or infinite elements of {xn}. In former case, choose a1 = a0 and b1 = a0+b02 and in later case

choose a1 = a0+b02 , b1 = b0. Then the interval [a1, b1] contains infinite elements of {xn} and b1 is still the

upper bound of the set {xn}. Continuously to subdivide the interval [ak−1, bk−1] into two subintervals and

choose one of these two intervals to be [ak, bk] which has the properties:

(1) the interval [ak, bk] contains infinite elements of {xn}

(2) bk is still the upper bound of the set {xn}


By this way we get a sequence of intervals {Ik}, where Ik = [ak, bk], and satisfy:

(1) ak ≤ ak+1 < bk+1 ≤ bk for all k

(2) bk − ak =b0 − a0

2k, k = 0, 1, · · ·

(3) every bk is an upper bound of {xn}.

We now show that both the sequences {ak} and {bk} are Cauchy sequences. For all ε > 0, we can choose

K such thatb0 − a0

2k< ε, ∀k ≥ K.

For this K , if k′ > k ≥ K , due to

ak ≤ ak′ ≤ bk,

we get

(ak′ − ak) ≤ bk − ak =b0 − a0

2k< ε,

i.e.

|ak′ − ak| < ε, for k′ > k ≥ K.

Similarly

(bk − bk′) ≤ bk − ak =b0 − a0

2k< ε,

i.e.

|bk′ − bk| < ε, for k′ > k ≥ K.

By definition, {ak} and {bk} are Cauchy Sequences, and both converge. From the theorem about operations

on convergent sequences, we obtain

limk→∞

bk − limk→∞

ak = limk→∞

(bk − ak) = limk→∞

b0 − a02k

= 0,

i.e., limk→∞

ak = limk→∞

bk. Let number C be the limit of both sequences. We show that C is the least upper

bound of {xn}. Because every bk is an upper bound of {xn}, that means for every fixed xn, the inequality

xn < bk, for all k

holds. As k → ∞, limk→∞

bk = C, hence for every xn

xn ≤ C,

which means that C is an upper bound of {xn}. Since limk→∞

ak = C, for any ε > 0, there must be some N

such that

aN > C − ε.

By the construction of the interval sequences {Ik}, the interval IN = [aN , bN ] contains infinite elements

of {xn}, hence such elements are larger than aN . Hence, the inequality aN > C − ε implies (C − ε) is no

longer an upper bound of {xn}. Therefore, C is the least upper bound of {xn}.


2.3 Exercise

1. Give a direct ′′ε−N ′′ verification of the following limits:

(a) limn→∞

1

n+ 5= 0.

(b) limn→∞

(2√n+

1

n+ 3

)= 3.

2. Suppose that the sequence {an} converges to a and that a > 0. Show that there is an index N such

that an > 0 for all indices n ≥ N .

3. Prove that the sequence {cn} converges to c if and only if the sequence {cn − c} converges to 0.

4. Prove that the Archimedean Property of R is equivalent to the fact that limn→∞

1

n= 0.

5. Show that if {an} converges, then limn→∞(an+1 − an) = 0.

6. Show that if limn→∞ bn = b < 0, then there is a natural numberN such that bn <

b

2∀n ≥ N.

7. Let an =1

n+ (−1)n. Show that lim

n→∞ an does not exist.

8. Consider the sequence xn =2

n3, n = 1, 2, · · · ,

(a) Let εk = 10−k with k = 1, 2, 3. Find Nk such that |xn − 0| < εk, ∀n ≥ Nk.

(b) For any ε > 0, find N such that |xn − 0| < ε, ∀n ≥ N.

9. Consider the sequence xn = 3√n, n = 1, 2, · · · ,

(a) Let Mk = 10k with k = 1, 2, 3. Find Nk such that xn > Mk, ∀n ≥ Nk.

(b) For anyM > 0, find N such that xn > M, ∀n ≥ N.

10. Assume that limn→∞ an = a.

(a) Prove that limn→∞ |an| = |a|.

(b) If an ≥ 0 and a ≥ 0, prove that limn→∞

√an =

√a.

11. Prove that limn→∞

(n2 +

√n− 1

n− 1000

)= ∞.

12. Discuss the convergence of each of the following sequences:

(a) limn→∞(

√n+ 1−√

n).

(b) limn→∞[(n+ 1)1/3 − n1/3].

13. Find the limit:

(a) limn→∞

3√n2 sinn!

n+ 1.


(b) limn→∞

[12

n3+

22

n3+ · · ·+ (n− 1)2

n3

].

(c) limn→∞

1 + a+ a2 + · · ·+ an

1 + b+ b2 + · · ·+ bn, (|a| < 1, |b| < 1).

(d) limn→∞

[12

n3+

32

n3+ · · ·+ (2n− 1)2

n3

].

14. Prove the following:

(a) limn→∞

n

2n= 0.

(b) limn→∞

5n

n!= 0.

(c) limn→∞

n!

nn= 0. Hint. n!

nn ≤ (1/2)n/2 .

(d) limn→∞

nk

n!, where k is a positive integer.

(e) limn→∞

n√n = 1. Hint. Define αn = n1/n − 1, then n = (1 + αn)

n ≥ 1 +n(n− 1)

2α2n.

15. Let Sn =1

2 · 1 +1

3 · 2 + · · ·+ 1

(n+ 1) · n, n ≥ 1. Prove that limn→∞Sn = 1.

16. Suppose that limn→∞ an = a. Define Gn =

a1 + a2 + · · ·+ ann

. Prove that limn→∞Gn = a.

Hint. We may assume a = 0, otherwise, let bn = an − a. ∃N1, s.t. |an| < ε/2, ∀n ≥ N1. Then∣∣∣ 1n

n∑i=1

ai

∣∣∣ ≤∣∣∣ 1n

N1∑i=1

ai

∣∣∣+∣∣∣ 1n

n∑i=N1+1

ai

∣∣∣ ≤∣∣∣ 1n

N1∑i=1

ai

∣∣∣+ ε

2. ∃N2, s.t.

∣∣∣ 1n

N1∑i=1

ai

∣∣∣ ≤ ε

2.

17. Suppose that 0 < a < 1.

(a) Prove that limn→∞ an = 0.

(b) If limn→∞ an = a, prove that lim

n→∞ ann = 0.

(c) Prove that limn→∞nan = 0.


(a) limn→∞

nk

an= 0 (a > 1). Hint. (a1/k)n = (1 + δ)n ≥ n(n− 1)δ2/2, then n/(a1/k)n → 0.

(b) limn→∞

n√a = 1 (a > 0). Hint. Let xn = n

√a− 1, then a = (xn + 1)n ≥ nxn, or, xn ≤ a/n → 0.

19. Show that limn→∞

(12· 34· · · 2n− 1

2n

)= 0.

20. Show that if limn→∞xn = a, and xn > 0, n = 1, 2, · · · , then lim

n→∞n√x1x2 · · ·xn = a.

21. Show that if limn→∞

xn+1

xn= a, and xn > 0, n = 1, 2, · · · , then lim

n→∞n√xn = a.

Hint. Let an = xn+1/xn, then an → a. n√xn = n

√x1

n√a1 · a2 · · · an/ n

√an. Let sn = n

√an − 1, then

an = (1 + sn)n ≥ nsn, or sn ≤ an/n → 0.


[1 · 3 · · · (2n− 1)

2 · 4 · · · 2n]1/n

= 1.



nn√n!

= limn→∞

(1 +

1

n

)n= e.

24. Show that the set (−∞, 0] is closed.

25. Show that b is the greatest lower bound of S if and only if b is a lower bound of S and ∀ε > 0, ∃x 0 ∈S such that x0 < b+ ε.

26. Prove that for any irrational number x, there exists a rational number sequence {r n} such that

limn→∞ rn = x.

27. Suppose that the sequence {an} is monotone. Prove that {an} converges if and only if {a2n} con-

verges. Show that this result does not hold without the monotonicity assumption.

28. Show that if the sequence {xn} converges, then any its subsequence {xpn} also converges and

converges to the same limit: limn→∞xpn = lim

n→∞ xn.

29. Show that if a subsequence of a monotonic sequence converges, then the monotonic sequence itself

converges.

30. Show that if limn→∞xn = ∞, then there exists a k such that xk = inf{xn}.

31. Prove the convergence of the following sequences by using the Monotone Convergence Theorem:

(a) xn = p0 +p110

+ · · ·+ pn10n

, n = 1, 2, · · ·, where pi, i = 0, 1, 2, · · · , are nonnegative integers,

and pi ≤ 9, i = 1, 2, · · · .(b) xn =

(1− 1

2

)(1− 1

4

)· · ·(1− 1

2n

).

(c) xn =10

1· 113

· · · n+ 9

2n− 1.

(d) x1 =√2, x2 =

√2 +

√2, x3 =

√2 +

√2 +

√2, · · · , xn =

√2 +

√2 + · · ·+√

2, · · ·.

32. Determine whether the following sequences converge or not by using Cauchy Convergence Criterion.

(a) xn = a1q + a2q2 + · · ·+ anq

n, where |q| < 1 and |ai| < M .

(b) xn = 1 +1

2+

1

3+ · · ·+ 1

n.

(c) xn =sin θ

2+

sin 2θ

22+ · · ·+ sinnθ

2n, where θ is a constant.

(d) xn = 1 +1

2+

1

22+ · · ·+ 1

2n.

(e) xn =cos 1!

1 · 2 +cos 2!

2 · 3 + · · ·+ cosn!

n · (n+ 1).

33. Prove that the sequence xn = 1 +1

2!+ · · ·+ 1

n!converges.

34. Let c > 0 and x1 > 0. Prove that the sequence xn+1 =√c+ xn, n ≥ 1 converges and find the

limit.


35. Let Sn =

n∑k=1

(−1)k−1

k, n ≥ 1. Show that {Sn} converges.

36. Prove the inequality:1

n+ 1< ln

(1 +

1

n

)<

1

n, n = 1, 2, · · · .

37. The sequence {xn} is defined by

x1 = a, x2 = b, xn =xn−1 + xn−2

2, n = 3, 4, · · · .

Find limn→∞xn.

Hint. xn − xn−1 = − 12(xn−1 − xn−2), then, x3 − x2 = − 1

2(x2 − x1) = − 1

2(b − a),

x4 − x3 = (− 12)2(b− a), x5 − x4 = (− 1

2)3(b − a), · · · , xn − xn−1 = (− 1

2)n−2(b− a).

Then we have xn − x2 = (b− a)((−1/2) + (−1/2)2 + · · ·+(−1/2)n−2 = (b−a)(−1− (−1/2)n−2)/3,and xn → (a + 2b)/3.

38. Show that the sequence

xn = 1 +1

2+

1

3+ · · ·+ 1

n− lnn, n = 1, 2, · · ·

converges, and hence we have the formula

1 +1

2+

1

3+ · · ·+ 1

n= C + lnn+ εn,

where C = 0.577216 · · · is called the Euler Constant, and εn → 0, as n → ∞.

Hint. ln(1 +

1

n

)<

1

n. xn =

n∑i=1

1

n−

n∑i=2

(ln i− ln(i− 1)) >1

n> 0.

xn − xn+1 = ln(1 +

1

n

)− 1

n+ 1> 0.

39. Suppose that the sequences {xn} and {yn} are defined by

x1 = a, y1 = b, xn+1 =√xnyn, yn+1 =

xn + yn2

, n = 1, 2, · · · .

where 0 ≤ a ≤ b. Show that {xn} and {yn} have the same limit

μ(a, b) = limn→∞ xn = lim

n→∞ yn,

where μ(a, b) is the arithmetic-geometric mean of a and b.

40. Let {x2k+1} and {x2k} be subsequences of {xn} and suppose limk→∞

x2k+1 = limk→∞

x2k = l. Prove

that limn→∞xn = l.

41. Suppose that {an} is a sequence of nonnegative real numbers, and {an} has no converged subse-

quences. Prove that

limn→∞ an = +∞.

Chapter 3: Continuity 49

42. Suppose that the sequence {xn} is bounded. Prove that if limn→∞xn does not exist, then there exist

two subsequences {xmk} and {xnk

} of {xn} such that

limk→∞

xmk= a, lim

k→∞xnk

= b,

and a �= b.

Hint. Since {xn} is bounded, from Bolzano-Weierstrass Theorem {xn} has a converged subsequence xnk → b.Since lim

n→∞ xn �= b, then ∃ε0 > 0, s.t., for any N , ∃xn(N), s.t., |xn(N) − a| ≥ ε0. The sequence

xn(N), N = 1, 2, · · · , is bounded, then it has a converged subsequence xmk → a, and |xmk −a| ≥ ε0. Takethe limit, we get |b− a| ≥ ε0, i.e., a �= b.


Chapter 3

Continuity

3.1 Limit of Function

3.1.1 Limit

Limit point. A point p is called a limit point of the set S if every neighborhood of p contains a point q �= p

such that q ∈ S.

Definition 3.1.1 (Limit of a function) Let f be a function defined in the domainD and x 0 be a limit point

ofD. We say that the limit of function f is l as x tends to x0, provided that for any given ε > 0, there exists

a δ > 0, such that for x ∈ D,

|f(x)− l| < ε, ∀ 0 < |x− x0| < δ.

In this case, we denote

limx→x0

f(x) = l.

Examples 3.1.1 Prove that

limx→1

x2 − 1

−x2 + 3x− 2= 2.

Proof. ∣∣∣ x2 − 1

−x2 + 3x− 2− 2∣∣∣ = 3|x− 1|

|2− x| .

For given ε > 0, let δ1 = 1/2. Then, if |x− 1| < δ1, we have

|2− x| = |1 + 1− x| ≥ 1− |x− 1| > 1− δ1 =1

2,

51


and3|x− 1||2− x| < 6|x− 1|.

Let δ2 = ε/6, and δ = min{δ1, δ2}, then we have

∣∣∣ x2 − 1

−x2 + 3x− 2− 2∣∣∣ ≤ 6|x− 1| < ε, ∀|x− 1| < δ,

i.e., limx→1

x2 − 1

−x2 + 3x− 2= 2.

Left limit and right limit If x is restricted from left (right) hand side tending to x 0, i.e., x < x0(x > x0),

then the limit l is called left (right) limit, denoted by

limx→x0−0

f(x) = l

(lim

x→x0+0f(x) = l

)

It is easy to show that the limit exists if and only if both the left and right limits exist and equal.

Theorem 3.1.2 Let f be a function defined in D and x0 is a limit point of D. Then limx→x0

f(x) exists if and

only if limn→∞ f(xn) exists, where {xn} is an arbitrary sequence in D and xn → x0.

Proof. The necessary part: Assume limx→x0

f(x) = l, we want to show limn→∞ f(xn) = l, for any sequence

xn → x0. Since limx→x0

f(x) = l, then for given ε > 0, there exists a δ > 0, such that for x ∈ D,

|f(x)− l| < ε, if |x− x0| < δ.

Since xn → x0, then for δ > 0, there exists a N > 0, such that,

|xn − x0| < δ ∀n ≥ N.

Thus we have

|f(xn)− l| < ε, ∀n ≥ N,

i.e., limn→∞ f(xn) = l.

The sufficient part: Assume limn→∞ f(xn) = l, for some sequence xn → x0, and let yn → x0 be any se-

quence. Since both {xn} and {yn} can be subsequences of another sequence {zn} (e.g., take terms

from {xn} and {yn} in turn.), limn→∞ f(zn) exists, and f(zn) → l, then f(yn) → l. We want to show

limx→x0

f(x) = l. We argue by contradiction. Suppose the conclusion is not true, which means that for some

ε0 > 0, for any δ > 0 we have

|f(x)− l| ≥ ε0, for some x satisfying |x− x0| < δ.

Hence for δn = 1n there is a point an such that

|an − x0| < δn, but |f(an)− l| ≥ ε0.


Thus we obtain a sequence {an} in D that converges to x0 but {f(an)} does not converge to l. This

contradicts with the assumption.


limx→0

x sin1

x= 0.

Proof. Let {xn} be any sequence with xn �= 0 and limn→∞xn = 0. Since

0 ≤∣∣∣∣xn sin 1

xn

∣∣∣∣ ≤ |xn| ,

By Squeezing Principle

limn→∞

∣∣∣∣xn sin 1

xn

∣∣∣∣ = 0.

Then from Theorem 3.1.2 we get limx→0

x sin1

x= 0,


limx→∞

(1 +

1

x

)x

= e.

Proof. We have defined

limn→∞

(1 +

1

n

)n

= e,

where n is a natural number. Let {kn} be any subsequence of {n}, then

limn→∞

(1 +

1

kn

)kn

= e.

Now for an arbitrary sequence {xn}, xn → ∞, we prove

limn→∞

(1 +

1

xn

)xn

= e.

Assume all xn > 1, let kn = [xn], we have

1 +1

kn + 1< 1 +

1

xn< 1 +

1

kn.

Then we have (1 +

1

kn + 1

)kn

<

(1 +

1

xn

)xn

<

(1 +

1

kn

)kn+1

.

Since

limn→∞

(1 +

1

kn + 1

)kn

= limn→∞

(1 + 1

kn+1

)kn+1

(1 + 1

kn+1

) =e

1= e,

limn→∞

(1 +

1

kn

)kn+1

= limn→∞

(1 +

1

kn

)kn

·(1 +

1

kn

)= e · 1 = e.


By Squeezing Principle

limn→∞

(1 +

1

xn

)xn

= e.

Since {xn} is an arbitrary sequence such that xn → ∞, therefore

limx→∞

(1 +

1

x

)x

= e.

Examples 3.1.4 Let f(x) =√x, x ≥ 0. Show that

limx→0+0

√x = 0.

Proof. {xn} be any sequence with xn �= 0 and limn→∞xn = 0. Hence for any give ε > 0, let ε1 = ε2 and

chooseN such that xn < ε1, ∀n ≥ N , then we have

√xn <

√ε1 = ε, ∀n ≥ N.

The result follows from Theorem 3.1.2.

Examples 3.1.5 Prove that

limx→0

sinx

x= 1.

Proof. From figure 3.1.1, the areas of the triangle OCB, the sector OCB, and the triangle OCD satisfy

1

2|OC||AB| ≤ 1

2|OC|2|x| ≤ 1

2|OC||CD|, |x| < π

2.

x

O A

B

C

D

Figure 3.1.1

Let |OC| = 1, then |AB| = | sinx|, |CD| = | tanx|. Hence,

| sinx| ≤ |x| ≤ | tanx|, |x| < π

2.

Then we have

| cosx| = | sinx|| tanx| ≤

| sinx||x| ≤ 1, |x| < π

2.


Since cos2 x =√1− sin2 x and | sinx| < |x|, then

| cosx| ≥√1− |x|2, |x| ≤ 1.

Since sinx and x have a same sign wher |x| ≤ 1,

√1− |x|2 ≤ sinx

x≤ 1, |x| ≤ 1.

Let {xn} be any sequence satisfying xn �= 0 and limn→∞xn = 0. Then there is a natural numberN such that

|xn| ≤ 1, ∀n ≥ N.

Hence √1− |xn|2 ≤ sinxn

xn≤ 1, n ≥ N.

By Squeezing Principle,

limn→∞

sinxnxn

= 1.

Therefore,

limx→0

sinx

x= 1.

Definition 3.1.3 (Limit at infinity) Let f(x) be defined on (a,+∞). We say that f(x) converges to l as x

tends to +∞ if for given ε > 0, there exists a constantM such that

|f(x)− l| < ε, ∀x ≥M,

and denote by

limx→+∞ f(x) = l.

Similar to the limit of sequences, we can define other types of limit.

3.1.2 Operations and properties

Similar to the limit of sequences, we have the following properties for the limit of functions.

Theorem 3.1.4 Suppose that

limx→x0

f(x) = a, and limx→x0

g(x) = b.

Then

(i) limx→x0

[(f(x) ± g(x)] = limx→x0

f(x)± limx→x0

g(x) = a± b.

(ii) limx→x0

[f(x) · g(x)] = limx→x0

f(x) · limx→x0

g(x) = a · b.


(iii) If g(x) �= 0, ∀x, and b �= 0, then, limx→x0

f(x)

g(x)=

limx→x0

f(x)

limx→x0

g(x)=a

b.

Theorem 3.1.5 (Squeezing Principle) Suppose that f(x) ≤ g(x) ≤ h(x), and

limx→x0

f(x) = limx→x0

h(x) = a.

then limx→x0

g(x) = a.

Examples 3.1.6 Let k be a positive integer, prove that

limx→+∞

xk

ax= 0 (a > 1).

Proof. We may assume that x > 1, then

0 ≤ xk

ax≤ ([x] + 1)k

a[x].

Since limn→∞

nk

an= 0, then

limx→+∞

([x] + 1)k

a[x]= lim

n→∞(n+ 1)k

an= 0.

By Squeezing Principle, limx→+∞

xk

ax= 0.

Definition 3.1.6 The function f : D → R is called monotonically increasing provided that

f(v) ≥ f(u), ∀u, v ∈ D s.t. v > u.

The function f : D → R is called monotonically decreasing provided that

f(v) ≤ f(u), ∀u, v ∈ D s.t. v > u.

In both cases, if the equality does not hold, then the function is called strictly monotone

(strictly increasing or strictly decreasing).

Theorem 3.1.7 (Monotone Convergence Theorem) Suppose that f(x) is defined on some interval (a, b)

and monotonically increasing. If f(x) is bounded above, i.e.,

f(x) ≤M ∀x ∈ (a, b),

then limx→b−

f(x) exists.

Theorem 3.1.8 (Cauchy Convergence Criterion) limx→x0

f(x) exists if and only if for given ε > 0, ∃δ > 0,

such that for any x and x′ satisfying 0 < |x− x0| < δ and 0 < |x′ − x0| < δ, we have

|f(x)− f(x′)| < ε.


Proof. Suppose limx→x0

f(x) = l, then for given ε > 0, ∃δ > 0, s.t.,

|f(x)− l| < ε

2, ∀0 < |x− x0| < δ,

|f(x′)− l| < ε

2, ∀0 < |x′ − x0| < δ.

Therefore,

|f(x)− f(x′)| ≤ |f(x)− l|+ |f(x′)− l| < ε. ∀0 < |x− x0| < δ, 0 < |x′ − x0| < δ.

On the other hand, let xn → x0 be any sequence. By assumption, for given ε > 0, ∃δ > 0, s.t.,

|f(x)− f(x′)| < ε, ∀0 < |x− x0| < δ, 0 < |x′ − x0| < δ.

For this δ > 0, since xn → x0, ∃N , s.t.,

0 < |xm − x0| < δ, 0 < |xn − x0| < δ ∀m,n ≥ N.

Then,

|f(xm)− f(xn)| < ε, ∀m,n ≥ N,

i.e., {f(xn)} is a Cauchy sequence, and converges. By Theorem 3.1.2, limx→x0

f(x) exists.

3.2 Continuity

3.2.1 Definition

Let f be a function defined in the domainD and x0 ∈ D is a limit point ofD.

Definition 3.2.1 (Continuity) The function f : D → R is said to be continuous at the point x 0 ∈ D

provided that

limx→x0

f(x) = f(x0).

In terms of ε− δ: For any given ε > 0, there exists a δ > 0, such that for x ∈ D,

|f(x)− f(x0)| < ε, ∀|x− x0| < δ.

The function f : D → R is said to be continuous in D provided that it is continuous at every point in D.

3.2.2 Operations and composition on continuous functions

Theorem 3.2.2 Suppose that the functions f : D → R and g : D → R are continuous in D. Then

(i) f(x) ± g(x) is continuous in D;


(ii) f(x) · g(x) is continuous in D;

(iii) if g(x) �= 0 in D, thenf(x)

g(x)is continuous in D.

Theorem 3.2.3 (Continuity of composition) Suppose that f : D → R is continuous at the point x 0 ∈ D

and g : U → R is continuous at the point f(x0) ∈ f(D) ⊆ U . Then the composition g ◦ f is continuous

at x0.

Proof. Let {xn} be a sequence in D and xn → x0. By the continuity of f at point x0, we get f(xn) →f(x0). Then {f(xn)} is a sequence in U and converges to f(x0), so by the continuity of g at point f(x0)

we get

g(f(xn)) → g(f(x0)),

which means that g ◦ f is continuous at x0.

3.2.3 Examples

Examples 3.2.1 The function

Pn(x) =

n∑k=0

akxn−k

is continuous everywhere.

Proof. It is obvious that f(x) = x is continuous in (−∞,∞). Hence, by the theorem of operations on

continuous functions, Pn(x) is continuous in (−∞,∞).

Examples 3.2.2 f(x) = sinx is continuous everywhere.

Proof. For any x0 ∈ (−∞,∞)

sinx− sinx0 = 2 sinx− x0

2cos

x− x02

.

It has been proved that | sinx| ≤ |x|, then

| sinx− sinx0| ≤ 2

∣∣∣∣sin x− x02

∣∣∣∣ ≤ 2|x− x0|

2= |x− x0|.

Hence, ∀ε > 0, choose δ = ε, then

| sinx− sinx0| ≤ |x− x0| < ε, if |x− x0| < δ,

which means that f(x) is continuous at x0.

Examples 3.2.3 f(x) =√x is continuous in [0,∞).

Proof. It has been proved f(x) is continuous at the point 0. For x 0 > 0, we have

|√x−√x0| = |x− x0|

|√x+√x0| ≤

|x− x0|√x0

.


∀ε > 0, choose δ =√x0ε, then

|√x−√x0| < ε, if |x− x0| < δ,

which means that f(x) is continuous at x0.

Examples 3.2.4 cosx is continuous everywhere; tanx is continuous in(−π

2 ,π2

).

Proof. Proof: Let f(x) = 1− sin2 x and g(x) =√x. f(x) is continuous in (−∞,∞), g(x) is continuous

in [0,∞), and f((−∞,∞)) ⊆ D(g). Since

cosx =

⎧⎨⎩√

1− sin2 x = g ◦ f(x), x ∈ [−π2 + 2kπ, π2 + 2kπ

]−√1− sin2 x = −g ◦ f(x), x ∈ [π2 + 2kπ, 3π2 + 2kπ

],

k = 0, 1, · · · . By the theorem of composition on continuous functions, cosx is continuous everywhere.

Since

tanx =sinx

cosx, x ∈

(−π2,π

2

),

it is continuous in(−π

2 ,π2

)due to the continuity of sinx and cosx.

Examples 3.2.5 The function

f(x) =

⎧⎨⎩

sin xx , x �= 0,

1, x = 0

is continuous everywhere.

Proof. By the theorem of operations on continuous functions, sin xx is continuous everywhere except x = 0.

We have proved that

limx→0

sinx

x= 1.

Since f(0) = 1, we have

limx→0

f(x) = f(0),

i.e., f(x) is continuous at x = 0.

Examples 3.2.6 Let

f(x) = limn→∞

x2n − 1

x2n + 1,

then f(x) is not continuous (discontinuous) at x = ±1.

Examples 3.2.7 Let

f(x) =

⎧⎨⎩

1x3 , x �= 0,

0, x = 0,

then f(x) is not continuous at x = 0.


Examples 3.2.8 Let X : R → R (Dirichlet function) be defined by

X (x) =

⎧⎨⎩ 1, x is rational,

0, x is irrational,

then X (x) is not continuous everywhere.

Proof. In case x0 is a rational number, X (x0) = 1. Let ε0 = 12 . By the density theorem, no matter what the

δ is chosen, the interval (x0 − δ, x0 + δ) always contains an irrational number x. Since X (x) = 0, we have

|X (x)− X (x0) = 1 > ε0,

i.e., X (x) is not continuous at x0. Similarly, we can prove the case of an irrational number.

3.3 Extreme Value Theorem

In sections 2-4, we will apply the theory mentioned in the first two chapters to prove three important

theorems about continuous functions defined in a bounded closed interval.

3.3.1 Maximizer and minimizer

Definition 3.3.1 We say that a function f : D → R attains a maximum (minimum) provided that its image

f(D) has a maximum (minimum); that is, there is a point x0 (not necessary only one) in D such that

f(x) ≤ f(x0) (f(x) ≥ f(x0)), ∀x ∈ D.

We call such a point in D a maximizer (minimizer) of the function f .

In general, no assertion can be made concerning the existence of a maximum or minimum. But we have

the following important theorem about a continuous function defined on a bounded closed interval [a, b].

3.3.2 Extreme value theorem

Theorem 3.3.2 (Extreme Value Theorem) Suppose that f : [a, b] → R is continuous. Then f attains both

a minimum and a maximum in [a, b].

We first to prove a weaker result as the following lemma.

Lemma 3.3.3 Suppose that f : [a, b] → R is continuous. Then the image of f : [a, b] → R is bounded;

that is, there is a number M such that

|f(x)| ≤M, ∀x ∈ [a, b].


Proof. We will argue by contradiction. If f([a, b]) is not bounded above, then for every natural number n,

there must be a point xn such that

f(xn) > n, n = 1, 2, 3, · · ·

Since a ≤ xn ≤ b, by Bolzano-Weierstrass Theorem, there is a subsequence {xnk} of {xn} converges to a

point x0 ∈ [a, b]. Hence on one hand, f(xnk) > nk i.e. {f(xn)} has no upper bound; on the other hand,

due to continuity of f, f(xnk) → f(x0).

This contradiction show that f([a, b]) is bounded above. Similarly we can prove that f([a, b]) is bound-

ed below. If it is not true, then we may choose a sequence {xn} such that

f(xn) < −n, n = 1, 2, 3, · · ·

Then we have a convergent subsequence {xnk} that on one hand {f(xn)} has no lower bound and on the

other hand f(xnk) → f(x0) as xnk

→ x0.

Proof of Theorem 3.3.2 Since f([a, b]) is bounded above, according to the Completeness Axiom, f([a, b])

has the least upper bound. Define C = sup f([a, b]). Now, we want to find a point x 0 ∈ [a, b] such that

f(x0) = C. Then

f(x0) = max f [a, b],

or,

f(x0) ≥ f(x), ∀x ∈ [a, b].

SinceC is the least upper bound of f([a, b]), for any natural number n, C− 1n is not an upper bound. Hence

there is a sequence of points xn at which

f(xn) > C − 1

n, n = 1, 2, 3, · · ·

Then we have

C − 1

n< f(xn) ≤ C,

it follows that the sequence {f(xn)} converges to C. Again according to Bolzano-Weierstrass theorem,

there is a subsequence {xnk} of {xn} that converges to a point x0 ∈ [a, b]. Since f is continuous on [a, b],

{f(xnk)} converges to f(x0). But {f(xnk

)} is a subsequence of {f(xn)}, so the limit of {f(xn)} is also

the limit of {f(xnk)}, i.e. f(x0) = C. Thus we have proved that x0 is a maximizer and f(x0) attains the

maximum.

Because f([a, b]) is bounded below, f([a, b]) has a greatest low bound. In a similarly way by Bolzano-

Weierstrass theorem we can find a minimizer at which f attains a minimum.


Note: If a continuous function defined in an open interval. It may not have maximum and minimum. The

reason is that if limn→∞xn = x0 and a < xn < b, the limit x0 may be the end point a or b.

Example 3.3.1 Let f(x) = x, 0 < x < 1. The function f is continuous in (0, 1), but it has no maximum

and minimum in the interval.

3.3.3 Compactness

Definition 3.3.4 A set S is said to be compact (sequentially compact) provided that every sequence in S has

a subsequence that converges to a point of S.

In this new terminology, the Bolzano-Weierstrass Theorem is simply the assertion that the set of a

bounded closed interval [a, b] is compact.

Theorem 3.3.5 Suppose that S is a compact nonempty set of real numbers and f : K → R is continuous.

Then f : S → R attains both a minimum and a maximum.

3.4 Intermediate Value Theorem

Theorem 3.4.1 Suppose that the function f : [a, b] → R is continuous, and

f(a) < 0, f(b) > 0.

Then there is at least a point x0 in the open interval (a, b) at which f(x0) = 0.

Proof. Let a1 = a, b1 = b and c1 = 12 (a1 + b1).

If f(c1) ≤ 0, define a2 = c1 and b2 = b1.

If f(c1) > 0, define a2 = a1 and b2 = c1.

In general, let cn = 12 (an + bn).

If f(cn) ≤ 0, define an+1 = cn and bn+1 = bn.

If f(cn) > 0, define an+1 = an and bn+1 = cn.

Hence we define a sequence of nested, closed intervals {[an, bn]} that satisfies f(an) ≤ 0, f(bn) > 0,

a ≤ an ≤ an+1 < bn+1 ≤ bn ≤ b,

and

bn − an =1

2(bn−1 − an−1) = · · · = 1

2n−1(b− a).


Thus the sequences {an} and {bn} satisfy the assumptions of the Nested Interval Theorem, so there is a

point x0 ∈ [a, b] such that an → x0 and bn → x0. Since f(x) is continuous on [a, b],

limn→∞ f(an) = lim

n→∞ f(bn) = f(x0).

Since f(an) ≤ 0, ∀n, we have f(x0) ≤ 0, and on the other hand we have f(x0) ≥ 0 due to f(bn) ≥ 0, ∀n.

Hence f(x0) = 0.

Theorem 3.4.2 (Intermediate Value Theorem) Suppose that the function f : [a, b] → R is continuous

and c is a number strictly between f(a) and f(b), i.e.,

f(a) < c < f(b), or f(a) > c > f(b).

Then there is a point x0 in the open interval (a, b) at which f(x0) = c.

Proof. First let us suppose that f(a) < c < f(b). Define the function g(x) = f(x)− c. Then

g(a) < 0 < g(b).

It follows from Theorem 3.3.1 that there is a point x0 ∈ (a, b) at which g(x0) = 0, i.e.

g(x0) = f(x0)− c = 0, or f(x0) = c.

In the case when f(a) > c > f(b), define g(x) = c− f(x) and follow the same argument.

Example 3.4.1 Let the integer k > 0 be odd, then the equation

akxk + ak−1x

k−1 + · · ·+ a0 = 0

has at least one real root.

Proof. We may assume that ak > 0. Let

f(x) = akxk + ak−1x

k−1 + · · ·+ a0.

Then, limx→+∞ f(x) = +∞, lim

x→−∞ f(x) = −∞. There exists M > 0, s.t., f(−M) < 0, f(M) > 0. By

Intermediate Value Theorem, there is a point x0 ∈ (−M,M), s.t., f(x0) = 0.

3.5 Uniform Continuity

Suppose that a function f : D → R is continuous inD. According to the ε− δ definition of continuity,

for x0 ∈ D and given ε > 0, there exists a δ > 0, such that for x ∈ D,

|f(x)− f(x0)| < ε, ∀|x− x0| < δ.


In general, δ depends not only on ε, but also on x 0, i.e., at different points, we need to choose different δ

such that the above inequality is true. Sometimes the δ may need to tend to 0 even for a fixed ε, for example,

f(x) = 1/x, x ∈ (0, 1). Then we may ask the question: What extra conditions are needed in order to find

a δ which is good for any x0 ∈ D?

Definition 3.5.1 (Uniformly continuity) Suppose that the function f(x) is defined on an interval D. f(x)

is said to be uniformly continuous in D provided that for any given ε > 0, ∃δ > 0, s.t.

|f(u)− f(v)| < ε, ∀u, v ∈ D, |u− v| < δ.

If D is an open interval, f : D → R is continuous in D, but f may not be uniformly continuous in D.

Example 3.5.1 Let f(x) = 1x , D = (0, 1), then f(x) is continuous in D, but not uniformly continuous in

D. Since for any u, v ∈ D, ∣∣∣∣ 1u − 1

v

∣∣∣∣ =∣∣∣∣u− v

uv

∣∣∣∣ .Take ε0 = 1, then for any δ > 0, let 0 < u0 < δ and v0 = u0

2 , we have |u0 − v0| = u0

2 < δ, but

|f(u0)− f(v0)| =∣∣∣∣ 1u0 − 1

v0

∣∣∣∣ = 1

u0> 1 = ε0.

If D is a bounded closed interval, we have the following theorem.

Theorem 3.5.2 A continuous function on a closed bounded interval, f : [a, b] → R, is uniformly continu-

ous.

Proof. We will argue by contradiction. Let D = [a, b]. Suppose that f : D → R is not uniformly

continuous. Then there is some ε0 > 0 such that ∀δ > 0

|f(u)− f(v)| ≥ ε0, for some u, v ∈ D, |u− v| < δ.

Let δ = 1n , then there exist points un and vn in [a, b] such that |un − vn| < 1

n , but

|f(un)− f(vn)| ≥ ε0, ∀n.

This defines two sequences {un} and {vn} in [a, b]. By using Bolzano-Weierstrass Theorem, we can choose

a subsequence {unk} of {un} that converges to a point u∗ ∈ [a, b]. But for each natural number k,

|unk− vnk

| < 1

nk≤ 1

k,

or

unk− 1

k< vnk

< unk+

1

k.

By Squeeze Principle, {vnk} also converges to u∗. The assumption of continuity of f at point u∗ implies

that both {f(unk)} and {f(vnk

)} converge to f(u∗). Thus their difference {f(unk) − f(vnk

)} converges


to 0. This contradicts with the inequality |f(unk)− f(vnk

)| ≥ ε0. It follows that f : D → R is uniformly

continuous.

3.6 Inverse Function

Definition 3.6.1 A function f : D → R is said to be one-to-one provided that for each point y in its image

f(D), there is exactly one point x in its domain D such that f(x) = y. In other words, if f(x1) = f(x2),

then x1 = x2. For a one-to-one function f : D → R, we define its inverse, denoted by

f−1 : f(D) → R,

as following:

f−1(y) = x, if f(x) = y.

It is clear that if f : D → R is strictly monotone, it is one-to-one and its inverse f −1 : f(D) → R is

also strictly monotone. In fact, if f is strictly increasing, then f −1 is also strictly increasing; if f is strictly

decreasing, then f−1 is also strictly decreasing.

Theorem 3.6.2 Let D be an interval and suppose that the function f : D → R is monotone. If its image

f(D) is an interval, then the function f is continuous.

Proof. Since −f is increasing if f is decreasing, we only need to consider the case when f is increasing.

For any x0 ∈ D, it is easy to show that limx→x0−0

f(x) = f(x0 − 0) and limx→x0+0

f(x) = f(x0 + 0) exist.

Suppose that f(x) is not continuous at x0, for example, f(x0− 0) < f(x0). Since f(x) ≤ f(x0− 0) when

x < x0 and f(x) ≥ f(x0) when x > x0, then f(x) can not take the values between f(x0 − 0) and f(x0),

which contradicts with the assumption that f(D) is an interval.

Theorem 3.6.3 Let I be an interval and suppose that the function f : I → R is strictly monotone. If f(I)

is an interval, then the inverse function f−1 : f(I) → R is continuous.

Proof. Suppose that f is strictly increasing. Let D = f(I). Then the function f −1 : D → R is a

monotonically increasing function that has the interval I as its image. From Theorem 3.6.3, f −1 : D → R

is continuous.

Theorem 3.6.4 Let I be an interval and the function f : I → R is strictly monotone and continuous. Then

the inverse function f−1 : f(I) → R is continuous and the definition domain f(I) is an interval.

Proof. Let y1 = f(x1), y2 = f(x2) ∈ f(I), x1 < x2, and c be any number in between y1 and y2.

Then from the Intermediate Value Theorem there exits a point x 0 ∈ (x1, x2) such that f(x0) = c. Hence

c ∈ f(I), i.e., f(I) is an interval. Since f(I) is an interval, by Theorem 3.6.3. f −1 is continuous.


Example 3.6.2 Define f : [0,∞) → R, as y = f(x) = x2. The inverse x = f−1(y) =√y. We have

proved that the function√y is continuous in [0,∞) by the definition of continuity. Now, we apply Theorem

3.6.4 to prove this. Because f(x) is strictly increasing and continuous in [0,∞) and the image of f is

f([0,∞)) = [0,∞). Hence by the theorem, f−1(y) =√y is continuous in [0,∞).

Example 3.6.2 Inverse functions sin−1 x, cos−1 x and tan−1 x.

We have proved the continuity of sinx, cosx and tanx. Since sinx is strictly increasing in[−π

2 ,π2

]and the image is [−1, 1], hence its inverse sin−1 x is continuous in [−1, 1] and also strictly increasing.

Because cosx is strictly decreasing in [0, π] and the image is [−1, 1], hence its inverse cos−1 x is

continuous in [−1, 1] and also strictly decreasing.

Because tanx is strictly increasing in[−π

2 ,π2

]and the image is (−∞,∞). Hence its inverse tan−1 x

is continuous on (−∞,∞) and also strictly increasing.

3.7 Continuity of Elementary Functions

In section 1 and 6, we have seen the continuity of polynomialsPn(x), the trigonometric functions sinx,

cosx, and tanx and their inverses. Using the properties of the continuous functions, we can easily show

that the rational functions and other trigonometric functions are all continuous in their domains. In this

section, we discuss the continuity of some other elementary functions, such as power functions, logarithmic

functions, and exponential functions. Then from the theorems for operations on continuous functions, for

inverse functions, and for compositions, we know that any combinations and operations (including inverse

and composition) on these functions result in also continuous functions.

3.7.1 Power Function with Rational Exponential

• y = x1n , x ≥ 0, n is a natural number.

Since x = yn is a strictly increasing function in interval [0,∞) and its image is also interval [0,∞),

its inverse function y = x1n is continuous in [0,∞).

• y = xmn , x ≥ 0, m,n are natural numbers.

This function can be treated as a composite function of z = x1n with y = zm. Hence y = x

mn is

continuous in [0,∞).

3.7.2 Exponential Function

• y = ax, a > 0.

Function y = aγ has been defined when γ is rational. Let’s temporarily assume a > 1. We consider

f(γ) = aγ in the rational set Q. Then f(γ) has the following properties.


(i) aγ1+γ2 = aγ1 · aγ2

(ii) f(γ) is strictly increasing, i.e., aγ1 < aγ2 , if γ1 < γ2.

(iii) For any sequence {γn} in Q that converges to 0, the sequence {f(γn)} converges to a0 = 1.

(iv) If {γn} is a convergent sequence, {aγn} is a Cauchy sequence.

• Definition of ax. For any real number x, let {γn} be a rational sequence and converges to x. It is

known from the above property (iv) that {aγn} is a Cauchy sequence. We define

ax = limn→∞ aγn .

For this definition, we need to prove that for any other rational sequence {γ ′n} that converges to x,

limn→∞ aγ

′n = lim

n→∞ aγn .

Indeed

limn→∞ aγ

′n = lim

n→∞

(aγ

′n − aγn

)+ lim

n→∞ aγn

= limn→∞ aγn

(aγ

′n−γn − 1

)+ lim

n→∞ aγn

= limn→∞ aγn · lim

n→∞

(aγ

′n−γn − 1

)+ lim

n→∞ aγn

= limn→∞ aγn .

Hence, for a > 1 we have define ax on the whole real axis, i.e. the interval (−∞,∞).

• Properties of Function ax

(i) For any real numbers x1 and x2, ax1+x2 = ax1 · ax2 .

(ii) For any real number x, ax > 0.

(iii) y = ax (a > 1) is strictly increasing.

Lemma 3.6.1 y = ax is continuous at x = 0.

Proof. We need to show that for any real sequence {xn} that converges to 0, {axn} converges to a0 = 1.

By the density property of rational numbers, we can choose two rational sequences {γ n} and {γ ′n} that

satisfy

xn − 1

n< γn < xn < γ′n < xn +

1

n.

when a > 1, ax is strictly increasing, hence

aγn < axn < aγ′n .


Since γn → 0 and γ ′n → 0 then aγn → 1 and aγ′n → 1. By using Squeezing Principle

limxn→0

axn = 1.

Up to now, we have assumed that a > 1. If 0 < a < 1, we define

ax = (a−1)−x,

and if a = 1, we define

ax = 1.

Obviously, y = ax is strictly decreasing, continuous at x = 0, and satisfies (i), (ii).

Theorem 3.6.2 The exponential function y = ax (a > 0) is continuous on (−∞,∞) and strictly increasing

(decreasing) when a > 1 (a < 1). Its image is the interval (0,∞)

Proof. Proof For any point x0 ∈ (−∞,∞)

ax − ax0 = ax0(ax−x0 − 1).

Since limx→0

ax = 1, we get

limx→x0

(ax − ax0) = limy→0

ax0(ay − 1) = 0,

which means that ax is continuous at x = x0.

3.7.3 Logarithmic Function

y = loga x and y = lnx: It has been known that the function x = ay is a strictly monotone and

continuous function from (−∞,∞) to (0,∞). Now we define its inverse by y = log a x (y = lnx for

a = e), which is from (0,∞) to (−∞,∞). Since ax (ex) is strictly monotone, its inverse loga x (ln x) is

continuous and also strictly monotone on (0,∞).

Example 3.7.1 Show that limx→0

ln(1 + x)

x= 1.

Proof. By the definition of Logarithm,

1

xln(1 + x) = ln(1 + x)

1x .

Hence, let f(x) = (1 + x)1x and g(y) = ln y. Since

limx→0

(1 + x)1x = lim

y→∞

(1 +

1

y

)y

= e,

then

limx→0

g(f(x) = g(limx→0

f(x))= g(e) = ln e = 1.


3.7.4 Power Function

y = xα (α is a real number.) : This function can be defined as a composition of y = f(x) = α ln x

with g(y) = ey , i.e., (g ◦ f)(x) = g(f(x)) = eα ln x. Since lnx is continuous in (0,∞) and the image

is (−∞,∞), ex is continuous in (−∞,∞). Hence by the theorem of continuity on composite functions,

y = xα = eα lnx is continuous in (0,∞).

3.8 Exercise

1. Let f(x) be defined by

f(x) =

⎧⎨⎩ n, if x = m

n , wherem and n > 0 are integers and have no common factors besides 1,

0, if x is irrational.

Show that f(x) is finite for any given x, but not bounded, i.e., f(x) is unbounded in any interval

containing x.

2. Show that the function f(x) =1 + x2

1 + x4is bounded in (−∞,+∞).

3. Show that the function f(x) =1

xcos

1

xis unbounded in any interval containing 0, but f(x) does not

tend to ∞ as x→ 0.

4. Discuss the boundedness of the function f(x) = lnx sin2 π

xfor 0 < x < ε.

5. Show that the function f(x) =x

1 + xhas the inf f(x) = 0 and sup f(x) = 1 in [0,+∞).

6. Find the inf and sup of the following functions:

(a) f(x) = x2, x ∈ (−2, 5).

(b) f(x) =1

1 + x2, x ∈ (−∞,+∞).

(c) f(x) =2x

1 + x2, x ∈ (0,+∞).

(d) f(x) = x+1

x, x ∈ (0,+∞).

(e) f(x) = 2x, x ∈ (−1, 2).

(f) f(x) = sinx, x ∈ (0,+∞).

(g) f(x) = sinx+ cosx, x ∈ [0, 2π].

(h) f(x) = [x], x ∈ (0, 2), or x ∈ [0, 2].

(i) f(x) = x− [x], x ∈ [0, 1].

7. Let m[f ] and M [f ] denote the greatest lower bound and the smallest upper bound of f(x). Prove

that if f1(x) and f2(x) are functions defined on (a, b), then

m[f1 + f2] ≥ m[f1] +m[f2], M [f1 + f2] ≤M [f1] +M [f2].

Give examples for the cases of equality and strict inequality.

8. Let f(x) =√1− x2, |x| ≤ 1. Show that lim

x→0f(x) = 1.


9. Let f(x) =

⎧⎨⎩ 1, x is irrational,

1n , x is rational andx = m

n ,

where m and n are integers with no common factors other than 1. Prove that limx→1

f(x) does not

exist.

10. Let f(x) = x3, x ∈ R. Verify the ε− δ criterion for limx→x0

f(x) = x30 at

(a) x0 = 2, (b) x0 = 50, (c) any point x0.

11. Let f(x) =1

x, x �= 0. Verify the ε− δ criterion for lim

x→x0

f(x) =1

x0at

(a) x0 = 10, (b) x0 = 1/10, (c) any point x0 �= 0.

12. Give a direct ε− δ verification of limx→2

x2 = 4, Find δ for ε = 0.1, 0.01, 0.001.

13. Give a direct M − δ verification of limx→1

1

(1 − x)2= +∞, Find δ for M = 10, 100, 1000.

14. Prove that limx→0

|x|2x

= 0.

15. Let P (x) = a0xn + a1x

n−1 + · · · + an, where ai, i = 0, 1, · · · , n, are real numbers. Show that

limx→∞ |P (x)| = +∞.

16. Let R(x) =a0x

n + a1xn−1 + · · ·+ an

b0xm + b1xm−1 + · · ·+ bm,

where a0 �= 0, b0 �= 0. Show that limx→∞R(x) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

∞, if n > m,

a0

b0, if n = m,

0, if n < m.

17. Find the limits:

(a) limx→1

x2 − 1√x− 1

. (b) limx→2

x4 − 16

x− 2.

18. Find the limits: (m and n are natural numbers.)

(a) limx→0

(1 + x)5 − (1 + 5x)

x2 + x5.

(b) limx→0

(1 +mx)n − (1 + nx)m

x2.

(c) limx→∞

(x+ 1)(x2 + 1) · · · (xn + 1)

[(nx)n + 1](n+1)/2.

(d) limx→3

x2 − 5x+ 6

x2 − 8x+ 15.

(e) limx→2

x3 − 2x2 − 4x+ 8

x4 − 8x2 + 16.

(f) limx→1

x+ x2 + · · ·+ xn − n

x− 1.

(g) limx→1

√x− 1

x− 1.

(h) limx→1

xm − 1

xn − 1.

(i) limx→a

(xn − an)− nan−1(x− a)

(x− a)2.

(j) limx→1

xn+1 − (n+ 1)x+ n

(x− 1)2.

(k) limx→1

(m

1− xm− n

1− xn

).

(l) limn→∞

(13 + 23 · · ·+ n3

n3− n

4

).

(m) limn→∞

13 + 43 + 73 + · · ·+ (3n− 2)3

[1 + 4 + 7 + · · ·+ (3n− 2)]2.



(a) limx→∞

1

n

[(x+

a

n

)+(x+

2a

n

)+ · · ·+

(x+

(n− 1)a

n

)].

(b) limx→∞

1

n

[(x+

a

n

)2+(x+

2a

n

)2+ · · ·+

(x+

(n− 1)a

n

)2].


(a) limx→+∞

√x+√x+

√x√

x+ 1.

(b) limx→+∞

√x+ 3

√x+ 4

√x√

2x+ 1.

(c) limx→4

√1 + 2x− 3√x− 2

.

(d) limx→−8

√1− x− 3

2 + 3√x

.

(e) limx→a

√x−√

a+√x− a√

x2 − a2.

(f) limx→3

√x+ 13− 2

√x+ 1

x2 − 9.

21. Find the limits: (n is an integer.)

(a) limx→−2

3√x− 6 + 2

x3 + 8.

(b) limx→16

4√x− 2√x− 4

.

(c) limx→∞

√9 + 2x− 5

3√x− 2

.

(d) limx→0

n√1 + x− 1

x.

(e) limx→0

√1− 2x− x2 − (1 + x)

x.

(f) limx→0

3√8 + 3x− x2 − 2

x+ x2.

22. Find the limits: (m and n are integers.)

(a) limx→0

3√27 + x− 3

√27− x

x+ 23√x4

.

(b) limx→0

√1 + x−√

1− x3√1 + x− 3

√1− x

.

(c) limx→7

√x+ 2− 3

√x+ 20

4√x+ 9− 2

.

(d) limx→0

m√1 + αx n

√1 + βx− 1

x.

(d) limx→0

3√1 + x

3 − 4√1 + x

4

1−√1− x2

.

(e) limx→0

x2

5√1 + 5x− (1 + x)

.

(f) limx→0

m√1 + αx− n

√1 + βx

x.


(a) limx→0

1− cosx

x2. (b) lim

x→0

tanx− sinx

x3.

24. Prove that limx→+∞

ax

x= +∞ (a > 1).

25. (Monotone convergence theorem) Suppose that f(x) is defined on some interval (a, b) and mono-

tonically increasing. If f(x) is bounded above, i.e., f(x) ≤ M ∀x ∈ (a, b), prove that limx→b−

f(x)

exists.

26. Let P (x) = a1x+ a2x2 + · · ·+ anx

n andm is an integer. Show that limx→0

m√1 + P (x) − 1

x=a1m.

27. Use ε− δ criterion to prove the continuity of the following functions:


(a) f(x) = ax+ b, −∞ < x < +∞.

(b) f(x) = x2, −∞ < x < +∞.

(c) f(x) = x3, −∞ < x < +∞.

(d) f(x) =√x, 0 ≤ x < +∞.

(e) f(x) = 3√x, −∞ < x < +∞.

(f) f(x) = sinx, −∞ < x < +∞.

(g) f(x) = cosx, −∞ < x < +∞.

(h) f(x) = tan−1 x, −π2< x < +

π

2.

28. Define f(x) =√x, ∀x ≥ 0. Verify the ε− δ criterion for continuity at x = 4 and x = 100.

29. Define f(x) =

⎧⎨⎩ x+ 1, if x ≤ 3/4,

2, if x > 3/4.

Use the ε− δ criterion to show that f is not continuous at x = 3/4.

30. Define h(x) = 1/(1 + x2), ∀x. Verify the ε− δ criterion for continuity at each point x0.

31. A function f : D → R is called a Lipschitz function if there is a constant c ≥ 0 such that

|f(u)− f(v)| ≤ c|u− v| ∀u, v ∈ D. Prove that a Lipschitz function is continuous in D.

32. Let f : D → R be continuous. Prove that the function g(x) = |f(x)| is continuous in D.

33. Suppose that f(x) is continuous and f(x) ≥ 0 in D. Prove that the function g(x) =√f(x) is

continuous in D.

34. Let f(x) =

⎧⎨⎩ x2, x ≥ 0,

x, x < 0.Show that f : R → R is continuous.

35. Let g(x) =

⎧⎨⎩ x2, if x is rational,

−x2, if x is irrational.At what points is the function continuous? Justify your answer.

36. Suppose that the function f : R → R is continuous at the point x0 and that f(x0) > 0. Prove that

there is an interval I ≡ (x0−1/n, x0+1/n), where n is a natural number, such that f(x) > 0 ∀x ∈ I .

37. Suppose that the function g : R → R is continuous and that g(x) = 0 if x is rational. Prove that

g(x) = 0 ∀x ∈ R.

38. Let f(x) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

1 + x2, x > 0,

0, x = 0,

−(1 + x2), x < 0.

Show that f : R → R is not continuous, but it has a continuous inverse.

39. Prove that limx→+∞

loga x

x= 0.

40. Find the limit limn→+∞ cos

x

2cos

x

22· · · cos x

2n.

41. Suppose that the function f : R → R is continuous at the point x0. Prove that there is an interval

I ≡ (x0 − 1/n, x0 + 1/n), where n is a natural number, such that f(x) < n ∀x ∈ I .


42. Define the function h : [1, 2] → R as follows: h(x) = 0 if x ∈ [1, 2] is irrational; h(x) = 1/n if

x ∈ [1, 2] is rational and x = m/n, where m and n are natural numbers having no common factors

other than 1.

(a) Prove that h : [1, 2] → R fails to be continuous at each rational number in [1, 2].

(b) Prove that if ε > 0, then the set {x ∈ [1, 2] | h(x) > ε} has only a finite number of points.

(c) Use part (b) to prove that h : [1, 2] → R is continuous at each irrational number in [1, 2].

43. Discuss the continuity of the following functions:

(a) f(x) = |x|.

(b) f(x) =x2 − 4

x− 2if x �= 2, and f(2) = A.

(c) If x �= −1, f(x) =1

(1 + x)2, f(−1) arbrary.

(d) If x �= 0, f(x) =∣∣∣sinxx

∣∣∣, f(0) = 1.

(e) If x �= 0, f(x) =sinx

|x| , f(0) = 1.

(f) If x �= 0, f(x) = sin1

x, f(0) arbrary.


(a) If x �= 0, f(x) = x sin1

x, f(0) = 0.

(b) If x �= 0, f(x) = e−1x2 , f(0) = 0.

(c) If x �= 1, f(x) =1

1 + e1

x−1

, f(1) arbrary.

(d) If x �= 0, f(x) = x lnx2, f(0) = a.

(e) f(x) = [x].

(f) f(x) =√x− [

√x].

45. Find the discontinuous points of the following functions:

(a) f(x) =x

(1 + x)2.

(b) f(x) =1 + x

1 + x3.

(c) f(x) =x2 − 1

x3 − 3x+ 2.

(d) f(x) =1x − 1

x+11

x−1 − 1x

.

(e) f(x) =x

sinx.

(f) f(x) =

√1− cosπx

4− x2.

46. Find the discontinuous points of the following functions:

(a) f(x) = cos21

x.

(b) f(x) = sgn(sin

π

x

).

(c) f(x) = tan−1 1

x.

(d) f(x) =1

1− ex

1−x

.

(e) f(x) =√x tan−1 1

x.

(f) f(x) = ex+1x .

(g) f(x) =1

lnx.


(a) f(x) = sgn(sinx).

(b) f(x) = x− [x].

(c) f(x) = x[x].

(d) [x] sinπx.

(e) f(x) = x2 − [x2].

(f) f(x) =[ 1x

].


48. Suppose that the function f : R → R has the properties that (1) f(u+ v) = f(u)+ f(v) ∀u, v ∈ R,

(2) f(1) = m. Prove that (1) f(x) = mx for all rational number x. (2) if f is continuous in R, then

f(x) = mx ∀x ∈ R.

49. Prove that Pn(x) = xn + a1xn−1 + · · ·+ an = 0 has at least one real root if n is an odd number.

50. Let f : [0,∞) → R be continuous and satisfy (1) f(0) = 0, (2) f(x) ≥ √x ∀x ≥ 0. Show that for

c > 0, there is some x0 such that f(x0) = c.

51. Suppose that S is a nonempty set of real numbers that is not sequentially compact. Prove that either

(i) there is an unbounded sequence in S or (ii) there is a sequence in S that converges to a point

x0 /∈ S.

52. If a set S contains an unbounded sequence, show that the function f : S → R, defined by f(x) =

x, ∀x ∈ S, is continuous but unbounded. If a set S contains a sequence that converges to a point

x0 /∈ S, show that the function f : S → R, defined by f(x) = 1/|x− x0|, ∀x ∈ S, is continuous

but unbounded.

53. Show that if S is a nonempty subset ofR that fails to be sequentially compact, then there is a function

f : S → R that is continuous but unbounded.

54. Suppose that the function f : (a, b) → R is continuous and that limx→a+

f(x) and limx→b−

f(x) exist.

Prove that f(x) is bounded on (a, b).

55. Prove that there is a solution of the equation x9 + x2 + 4 = 0, x ∈ R.

56. Prove that there is a solution of the equation1√

x+ x2+ x2 − 2x = 0, x > 0.

57. Consider the equation x = f(x),−1 < x < 1. If f(x) is continuous in [−1, 1] and f(−1) >

−1, f(1) < 1, show that the equation has a solution in (−1, 1).

58. For a function f : D → R, a solution of the equation f(x) = x, x ∈ D, is called a fixed point of

f . If f : [−1, 1] → R is continuous, f(−1) > −1, and f(1) < 1, show that f : [−1, 1] → R has a

fixed point.

59. Suppose that the function h : [a, b] → R and g : [a, b] → R are continuous. Show that if

h(a) ≤ g(a) and h(b) ≥ g(b), then the equation h(x) = g(x), x ∈ [a, b], has a solution.

60. Suppose that f : R → R is continuous and its image f(R) is bounded. Prove that there is a solution

of the equation f(x) = x, x ∈ R.


61. Suppose that the function f(x) is continuous on [0, 1] and satisfies the conditions f(0) = f(1). Show

that for any natural number n, there exists at least one point ξ ∈ [0, 1] such that f(ξ) = f(ξ + 1

n

).

Hint. By contradiction. If it is not true, then ∃N > 1, s.t. f(x) �= f(x + 1N), ∀x ∈ [0, 1 − 1

N]. Let

g(x) = f(x) − f(x+ 1N), then g(x) > 0 or g(x) < 0, ∀x ∈ [0, 1 − 1

N], since g(x) is continuous. Suppose

g(x) > 0, then f(0) − f( 1

N

)> 0, f

( 1

N

) − f( 2

N

)> 0, · · · f(N − 1

N

) − f(1) > 0. Adding up we get

f(0) − f(1) > 0, which is a contradiction.

62. Suppose that f : [a, b] → R is continuous, and for every x ∈ [a, b], there exists y ∈ [a, b]

such that |f(y)| ≤ 12 |f(x)|. Prove that there exists a point ξ ∈ [a, b] such that f(ξ) = 0.

Hint. Let x1 = a, then |f(xn)| ≤ 1

2n−1|f(a)| → 0. ∃ subsequence {xnk} → ξ, f(xnk ) → f(ξ) = 0.

63. Suppose that f(x) is continuous in [a, b]. Let x1, x2, · · · , xn be points in [a, b]. Prove that there is a

point c ∈ [a, b] at which f(c) =f(x1) + f(x2) + · · ·+ f(xn)

n.

64. Suppose that the function f : [0, 1] → R is continuous and that its image consists entirely of rational

numbers. Prove that f(x) is a constant function.

65. Suppose that f : [0,+∞) → R is continuous. Let S be the set of real numbers defined by

S = {x | x = limn→+∞ f(xn) for some sequence {xn} ⊂ [0,+∞) such that lim

n→+∞xn = +∞.}

Prove that if a ∈ S and b ∈ S, then [a, b] ⊂ S.

Hint. Let c ∈ (a, b), and a ∈ S, b ∈ S. By assumption, ∃{xn} and {yn} s.t.,lim

n→+∞xn = +∞, limn→+∞ yn = +∞, and lim

n→+∞ f(xn) = a, limn→+∞ f(yn) = b.

Let ε =1

2min{(c− a), (b− c)}, then ∃N1, s.t., f(xn) < a+ ε < a+ 1

2(c+ a) < c, ∀n ≥ N1.

∃N2, s.t., f(yn) > b− ε > b− 12(b− c) > c, ∀n ≥ N2. Let N = max{N1, N2},

then f(xn) < c < f(yn), ∀n ≥ N . Since f(x) is continuous on (xn, yn), from Intermediate Value Theorem,∃zn ∈ (xn, yn), s.t., f(zn) = c. Therefore, c ∈ S, i.e., [a, b] ⊆ S.

66. Show that it is not necessary the case that if f : D → R and g : D → R are each uniformly

continuous, then so is the product fg : D → R.

67. Suppose that the functions f : D → R and g : D → R are uniformly continuous and bounded.

Prove that fg : D → R also is uniformly continuous.

68. A function f : D → R is called a Lipschitz function if there is a constant c ≥ 0 such that |f(u)−f(v)| ≤ c|u− v| ∀u, v ∈ D. Prove that a Lipschitz function is uniformly continuous in D.

69. Prove that each of the following functions is uniformly continuous in its definition domain by ε − δ

verification:

(a) f(x) =1

1 + x2, −∞ < x < +∞.

(b) f(x) = x2, −M < x < M, M > 0.

(c) f(x) =1

x, a ≤ x < +∞, a > 0.

70. Let f(x) =√x, x ∈ [0, 1]. Prove that f(x) is not a Lipschitz function but uniformly continuous in

[0, 1].


71. Prove that each of the following functions is not uniformly continuous in its definition domain:

(a) f(x) = x2, 0 ≤ x < +∞.

(b) f(x) =1

x, 0 < x < 1.

(c) f(x) = sin1

x, 0 < x < 1.

72. Suppose that a continuous function f : R → R is periodic, that is, there is number p > 0 such that

f(x+ p) = f(x) ∀x. Show that f : R → R is uniformly continuous.

73. Prove that the function f(x) = sinπ

xis continuous and bounded in (0, 1), but not uniformly contin-

uous.

74. Prove that the function f(x) = sinx2 is continuous and bounded in (−∞,+∞), but not uniformly

continuous.

75. Show that if f(x) is continuous in [a,+∞) and limx→+∞ f(x) exists, then f(x) is uniformly continuous

in [a,+∞).

76. Show that f(x) = x+ sinx is uniformly continuous in (−∞,+∞).

77. Discuss whether the following functions are uniformly continuous:

(a) f(x) =x

4− x2, −1 ≤ x ≤ 1.

(b) f(x) = lnx, 0 < x < 1.

(c) f(x) =sinx

x, 0 < x < π.

(d) f(x) = ex cos1

x, 0 < x < 1.

(e) f(x) = tan−1 x, −∞ < x < +∞.

(f) f(x) =√x, 1 ≤ x < +∞.

(g) f(x) = x sinx, 0 ≤ x < +∞.

78. Prove that the function f(x) =| sinx|x

is uniformly continuous in each of the intervals J1 = (−1, 0)

and J2 = (0, 1), but not uniformly continuous in the union J 1 ∪ J2 (0 < |x| < 1).

79. For given ε > 0, find the δ(ε), for which the function f(x) is uniformly continuous:

(a) f(x) = 5x− 3, −∞ < x < +∞.

(b) f(x) = x2 − 2x− 1, −2 ≤ x ≤ 5.

(c) f(x) =1

x, 0.1 ≤ x ≤ 1.

(e) f(x) =√x, 0 ≤ x < +∞.

(f) f(x) = 2 sinx− cosx, −∞ < x < +∞.

(g) f(x) = x sin1

x, x �= 0, and f(0) = 0, 0 ≤ x ≤ π.

80. Suppose that f(x) is continuous and monotone in a finite or infinite open interval (a, b). Prove that

f(x) is uniformly continuous in (a, b) if and only if f(x) is bounded in (a, b).

81. Let f(x) be continuous in a finite interval (a, b). Prove that f(x) can be continuously extended to

[a, b] if and only if f(x) is uniformly continuous in (a, b).

82. Show that f(x) = ax, −∞ < x < +∞, where a = f(1) is an arbitrary constant, is the unique

continuous function that satisfies the equation f(x+ y) = f(x) + f(y), ∀x, y ∈ R.

Chapter 4: Differentiation 77

83. Show that f(x) = ax, −∞ < x < +∞, where a = f(1) is a positive constant, is the unique contin-

uous function that is not identically zero and satisfies the equation f(x+y) = f(x)f(y), ∀x, y ∈ R.

84. Show that f(x) = loga x, x > 0, where a is a positive constant, is the unique continuous function

that is not identically zero and satisfies the equation f(xy) = f(x)f(y), ∀x, y > 0.

85. Show that f(x) = xa, x > 0, where a is a constant, is the unique continuous function that is not

identically zero and satisfies the equation f(xy) = f(x)f(y), ∀x, y > 0.

86. Find all continuous functions f(x), −∞ < x < +∞, that satisfies the equation f(x+ y) + f(x−y) = 2f(x)f(y), ∀x, y ∈ R.

87. Find all bounded continuous functions f(x) and g(x), −∞ < x < +∞, where f(0) = 1 and

g(0) = 0, that satisfies the equations f(x+ y) = f(x)f(y)− g(x)g(y) and g(x+ y) = f(x)g(y) +

f(y)g(x), ∀x, y ∈ R. Hint: Consider the function F (x) = f 2(x) + g2(x).

88. Let Δf(x) = f(x+Δx) − f(x) and Δ2f(x) = Δ(Δf(x)) denote the first and second order finite

difference of f(x). Show that if f(x), −∞ < x < +∞, is continuous and Δ2f(x) ≡ 0, then f(x)

is a linear function, i.e., f(x) = ax+ b, where a and b are constants.

89. Suppose that the function f : [0, 1] → R is continuous, f(0) > 0, and f(1) = 0. Prove that there is

a number x0 ∈ (0, 1] such that f(x0) = 0 and f(x) > 0 for 0 ≤ x ≤ x0; that is, there is a smallest

point in the interval [0, 1] at which the function f attains the value 0.

90. Let ωf (δ) = sup |f(x1) − f(x2)|, where x1, x2 ∈ (a, b) satisfying the condition |x1 − x2| ≤ δ.

Prove that f(x) is uniformly continuous in (a, b) if and only if limδ→+0

ωf (δ) = 0.

91. Suppose that f(x) is uniformly continuous in an open interval (a, b). Prove that f(x) is bounded in

(a, b).

92. Suppose that the function f : R → R is uniformly continuous and f(0) = 0. Prove that

|f(x)| ≤ 1 +M |x|, ∀x ∈ (−∞,+∞),

where M > 0 is a constant.

Hint. Suppose x ≥ 0. Since f(x) is uniformly continuous on [0, x], then for given ε = 1, ∃δ > 0, s.t., |f(x′)−f(x′′)| < ε, ∀|x′−x′′| < δ, x′, x′′ ∈ [0, x]. Divide [0, x] into subintervals [0, δ], [δ, 2δ], · · · , [kδ, x], wherekδ < x and x− kδ ≤ δ. Then we have|f(x)| = |f(x)− f(0)| ≤ |f(x)− f(kδ)|+ |f(kδ)− f((k − 1)δ)| + · · ·+ |f(δ) − f(0)|

≤ 1 + k = 1 +1

δkδ < 1 +

1

δx.

Let M = 1δ

, then |f(x)| ≤ 1 +M |x|. Similarly, if x < 0, then |f(x)| ≤ 1 +M |x|.


Chapter 4

Differentiation

4.1 Derivative

4.1.1 Definition

Definition 4.1.1 Let f(x) be a function defined in interval (a, b) and x0 ∈ (a, b). Consider the following

ratiof(x)− f(x0)

x− x0or

f(x0 +Δx) − f(x0)

Δx.

If the ratio has a limit l as x tends to x0 (or Δx → 0). Then we say that f(x) is differentiable at the point

x0 and l is the derivative of f at x0, denote by l = f ′(x0). Namely

limx→x0

f(x)− f(x0)

x− x0= f ′(x0),

or

limΔx→0

f(x0 +Δx) − f(x0)

Δx= f ′(x0).

The one side limits

limΔx→0−

f(x0 +Δx)− f(x0)

Δxand lim

Δx→0+

f(x0 +Δx)− f(x0)

Δx.

are called left and right derivative of f(x), and denoted by f ′−(x0) and f ′

+(x0).

Obviously, f(x) is differentiable at the x0 if and only if both f ′−(x0) and f ′+(x0) exist and equal.

By the definition, if f(x) is differentiable at x0, we have

f(x)− f(x0) = f ′(x0)(x − x0) + α(x)(x − x0),

where

α(x) =f(x)− f(x0)

x− x0− f ′(x0)

79


and satisfies

limx→x0

α(x) = 0.

Hence

limx→x0

[f(x)− f(x0)] = 0.

It means that we have the following proposition.

Proposition 4.1.2 If f(x) is differentiable at x0, then f(x) is continuous at x0.

The opposite assertion may not be true. Consider the following example:

Example 4.1.1 f(x) = |x| is continuous at x = 0 but not differentiable at x = 0.

Proof: Since

limx→0x>0

f(x)− f(0)

x− 0= lim

x→0x>0

x

x= 1,

limx→0x<0

f(x)− f(0)

x− 0= lim

x→0x<0

−xx

= −1.

Hence, the ratio f(x)−f(0)x−0 has no limit as x→ 0.

Example 4.1.2 f(x) = xn , f ′(x) = nxn−1

Proof: Using Binomial Formula we have

(x+Δx)n = xn + nxn−1Δx+

n∑j=2

(n

j

)xn−jΔxj ,

(x+Δx)n − xn

Δx= nxn−1 +

n∑j=2

(n

j

)xn−jΔxj−1.

Hence

limΔx→0

(x +Δx)n − xn

Δx= nxn−1.

Example 4.1.3 f(x) = sinx , f ′(x) = cosx

Proof: sin(x+Δx)− sinx = 2 sin Δx2 cos(x + Δx

2 ). Since cosx is continuous everywhere and

limΔx→0

sin Δx2

Δx2

= 1.

We have

f ′(x) = limΔx→0

f(x+Δx)− f(x)

Δx= lim

Δx→0

sinΔx2

Δx2

· cos(x+Δx

2)

= limΔx→0

sinΔx2

Δx2

· limΔx→0

cos(x+Δx

2) = cosx.


4.1.2 Physical and geometric interpretation

1. Velocity in Mechanics. Let S = S(t) be the function of displacement. Then S(t0+Δt)−S(t0)Δt is the

average velocity during the time period from t 0 to t0 +Δt, and

V (t0) = limΔt→0

S(t0 +Δt)− S(t0)

Δt= S′(t0)

is the instant velocity at time t0.

2. Slope of a tangent line.

The tangent line at point x0 is y = f(x0) +m0(x− x0) wherem0 = tan θ0, the slope of the line

m0 = limx→x0

f(x)− f(x0)

x− x0= lim

x→x0

tan θ = tan θ0 =⇒ Tangent Line:y = f(x0) + f ′(x0)(x − x0)

4.1.3 Algebra of Derivatives

Theorem 4.1.3 Suppose that f(x) and g(x) are differentiable at x0 ∈ (a, b). Then

1. (f + g)′(x0) = f ′(x0) + g′(x0).

2. (f · g)′(x0) = f(x0)g′(x0) + f ′(x0)g(x0).

3. if g(x) �= 0 for all x in (a, b), then (f/g)′(x0) =g(x0)f

′(x0)− f(x0)g′(x0)

[g(x0)]2.

Proof of 2: Since

f(x0 +Δx)g(x0 +Δx)− f(x0)g(x0) = f(x0 +Δx)g(x0 +Δx)

−f(x0 +Δx)g(x0) + f(x0 +Δx)g(x0)− f(x0)g(x0),

we have

f(x0 +Δx)g(x0 +Δx)− f(x0)g(x0)

Δx=

f(x0 +Δx)g(x0 +Δx)− g(x0)

Δx+f(x0 +Δx)− f(x0)

Δxg(x0)


Take limits on both sides as Δx→ 0 we get

(fg)′(x0) = f(x0)g′(x0) + f ′(x0)g(x0).

Proof of 3: Since

f(x0 +Δx)

g(x0 +Δx)− f(x0)

g(x0)=f(x0 +Δx)g(x0)− f(x0)g(x0 +Δx)

g(x0 +Δx)g(x0)

=[f(x0 +Δx)− f(x0)]g(x0)− f(x0)[g(x0 +Δx) − g(x0)]

g(x0 +Δx)g(x0)

Dividing both sides by Δx and taking limits we get

(f/g)′(x0) =g(x0)f

′(x0)− f(x0)g′(x0)

[g(x0)]2.

4.2 Differentiating Inverses and Compositions

4.2.1 Inverse

Theorem 4.2.1 Let I be an open interval containing the point x 0. Suppose that the function f : I → R is

strictly monotone and continuous, f is differentiable at x0, and f ′(x0) �= 0. Then the inverse f−1 : f(I) →R is differentiable at the point y0 = f(x0) and

(f−1)′(y0) =1

f ′(x0)

Proof. It follows from Theorem 3.6.4 that f(I) is an open interval containing the point y 0 = f(x0). For a

point y ∈ f(I), there is a point x ∈ I such that y = f(x) and x = f−1(y). Then we have

f−1(y)− f−1(y0)

y − y0=

1

f(x)− f(x0)

x− x0

.

Since the inverse function is continuous and monotone, then y → y 0 if and only if x → x0. Since f is

strictly monotone, f−1(y) �= f−1(y0) if and only if y �= y0. Hence,

limy→y0

f−1(y)− f−1(y0)

y − y0= lim

x→x0

1

f(x)− f(x0)

x− x0

=1

f ′(x0).

That is

(f−1)′(y0) =1

f ′(x0).

Corollary 4.2.2 Suppose that f is strictly monotone and differentiable with f ′(x) �= 0 in an open interval

I . Then the inverse function f−1 is differentiable in the interval f(I).


4.2.2 Composition and Chain Rule

Theorem 4.2.3 Let I be an open interval containing a point x 0 and suppose that the function f : I → R

is differentiable at x0. Let J be an open interval for which f(I) ⊆ J , and suppose that g : J → R is

differentiable at y0 = f(x0). Then the composition gof : I → R is differentiable at x0 and

(gof)′(x0) = g′(y0)f ′(x0).

Proof. Let y = f(x) with x �= x0. If there is an open interval containing x0 in which f(x) �= f(x0) if

x �= x0, then(gof)(x) − (gof)(x0)

x− x0=g(y)− g(y0)

y − y0

f(x)− f(x0)

x− x0.

By taking limits in both sides, we get the required result. To account for the possibility that no such interval

exists, we introduce an auxiliary function h : J → R by defining

h(y) =

⎧⎨⎩ [g(y)− g(y0)]/(y − y0), if y �= y0,

g′(y0), if y = y0.

Then

g(y)− g(y0) = h(y)(y − y0)

and(gof)(x)− (gof)(x0)

x− x0= h(y)

[f(x)− f(x0)

x− x0

].

From the definition of h(y), h is continuous at y0. Because differentiability implies continuity, f is contin-

uous at x0. Therefore, the composite function h(f(x)) is continuous at x0. Hence,

limx→x0

g(f(x))− g(f(x0))

x− x0= h(f(x0))f

′(x0) = g′(y0)f ′(x0).

The formula

(gof)′(x0) = g′(f(x0))f ′(x0)

is called the Chain Rule for differentiating composite function.

4.3 Derivatives of Elementary Function

1. f(x) = xn, f ′(x) = nxn−1, x ∈ (−∞,∞).

2. f(x) = ex, f ′(x) = ex, x ∈ (−∞,∞).

Proof. It’s known that limΔx→0

eΔx − 1

Δx= 1. Therefore,

limΔx→0

ex+Δx − ex

Δx= lim

Δx→0ex

eΔx − 1

Δx= ex.


3. f(x) = lnx, f ′(x) = 1x , x ∈ (0,∞).

Proof. It’s known that by definition x = f −1(y) = ey . By the property of differentiating inverse

f ′(x) =1

(f−1)′(y)=

1

ey=

1

x..

4. f(x) = ax, f ′(x) = ax · ln a, a > 0, x ∈ (−∞,∞).

Proof. f(x) = ax = ex ln a. Let y = g(x) = x ln a, and h(y) = ey . Then f(x) = h(g(x)) =

hog(x). By the property of differentiating composition f ′(x) = (hog)′(x) = h′(y)g′(x) = ey ·ln a = ex ln a · ln a = ax ln a.

5. f(x) = xα, f ′(x) = αxα−1, x ∈ (0,∞).

6. f(x) = sinx, f ′(x) = cosx, x ∈ (−∞,∞).

7. f(x) = cosx, f ′(x) = − sinx, x ∈ (−∞,∞).

8. f(x) = tanx, f ′(x) = 1

cos2 x= sec2 x, x ∈ (−π

2 ,π2 ).

Proof. Since tanx =sinx

cosx, we have

f ′(x) =cosx · cosx− sinx(− sinx)

cos2 x=

cos2 x+ sin2 x

cos2 x=

1

cos2 x.

9. (sin−1 x)′ = 1/√1− x2, x ∈ (−1, 1),

(cos−1 x)′ = −1/√1− x2, x ∈ (−1, 1),

(tan−1 x)′ = 11+x2 , x ∈ (−∞,∞).

4.4 Mean Value Theorems

Let f be a continuous function defined on (a, b). For a point x 0 ∈ (a, b), if there is a δ > 0 such that

f(x0) ≥ f(x), (or f(x0) ≤ f(x) ), for x0 − δ < x < x0 + δ,

then we say f(x0) is a local maximum (or minimum) and x0 is a local maximizer (or minimizer). In both

cases, f(x0) is called a local extreme.

Lemma 4.4.1 Let I be a neighborhood of x0 and suppose that the function f : I → R is differentiable at

x0. If the point x0 is either a local minimizer or local maximizer, then f ′(x0) = 0.

Proof. By the definition of derivative

limx→x0x<x0

f(x)− f(x0)

x− x0= lim

x→x0x>x0

f(x)− f(x0)

x− x0= f ′(x0).


First suppose that x0 is a local maximizer, then there is a δ > 0 such that

f(x)− f(x0)

x− x0≥ 0 for x0 − δ < x < x0

and hence

f ′(x0) = limx→x0x<x0

f(x)− f(x0)

x− x0≥ 0.

On the other hand,f(x)− f(x0)

x− x0≤ 0 for x0 < x < x0 + δ

and hence

f ′(x0) = limx→x0x>x0

f(x)− f(x0)

x− x0≤ 0.

Therefore

f ′(x0) = 0.

In the case when x0 is a local minimizer, the proof is similar.

x

y

a b

Theorem 4.4.2 (Rolle’s Theorem) Suppose that the function f : [a, b] → R is continuous and f is differ-

entiable in (a, b). Assume, moreover, that

f(a) = f(b).

Then there is a point x0 in (a, b) at which f ′(x0) = 0.

Proof. Since f(x) is continuous in [a, b], according to the Extreme Value Theorem, it attains both a mini-

mum value and a maximum value on [a, b]. Since f(a) = f(b), if both the maximizer and minimizer occur

at the end points, then f(x) is a constant, so f ′(x) = 0 in whole (a, b). Otherwise, the function has either a

maximizer or a minimizer at some point x0 ∈ (a, b). Hence, by Lemma 4.4.1, at this point f ′(x0) = 0.

Rolle’s Theorem is a special case of the following Lagrange Mean Value Theorem, but in fact, it has

some important applications by itself.

Theorem 4.4.3 (Lagrange Mean Value Theorem) Suppose that the function f(x) is continuous in [a, b]

and differentiable in (a, b). Then there is a point x0 ∈ (a, b) at which

f ′(x0) =f(b)− f(a)

b− a.


x

y

a b

b a-

f(b) f(a)-

( , ( ))x f x0 0

Proof. Let y = g(x) be the line passing through the points (a, f(a)) and (b, f(b)), namely

y = g(x) = f(a) +f(b)− f(a)

b− a(x− a).

Define the auxiliary function

d(x) = f(x)− g(x) = f(x)− f(a)− f(b)− f(a)

b− a(x− a).

Then d(x) is continuous on [a, b] and differentiable in (a, b), and d(a) = d(b) = 0. We apply Rolle’s

theorem to select a point x0 ∈ (a, b) at which d′(x0) = 0.

d′(x0) = f ′(x0)− f(b)− f(a)

b− a= 0,

or,

f ′(x0) =f(b)− f(a)

b− a.

Example 4.4.1 Show that if f ′(x) exists for all x ∈ (a, b), then f ′(x) cannot have jump discontinuity.

Proof. Let x0 ∈ (a, b), and take Δx > 0 such that x0 + Δx ∈ (a, b). Then, f(x) is continuous on

[x0, x0 +Δx], and differentiable on (x0, x0 +Δx). Using Lagrange Mean Value Theorem we get

f(x0 +Δx) − f(x0)

Δx= f ′(x0 + θΔx), 0 < θ < 1.

If f ′(x) has a jump discontinuity at x0, then limx→x0

f ′(x) exists. Then we have

f ′(x0) = f ′+(x0) = lim

Δx→0

f(x0 +Δx) − f(x0)

Δx= lim

Δx→0f ′(x0 + θΔx).

Similarly,

f ′(x0) = f ′−(x0) = lim

Δx→0f ′(x0 − θΔx),

i.e., f ′(x) is continuous at x0, which is a contradiction.

Example 4.4.2 Let f : (−1, 1) → R be defined by

f(x) =

⎧⎨⎩ x2 sin

(1x

), x �= 0,

0 x = 0.


Show that if f ′(x) exists for all x ∈ (−1, 1), but f ′(x) is not continuous at x = 0.

Proof. It is easy to see that f ′(x) exists for all x ∈ (−1, 1). If x �= 0, then

f ′(x) = 2x sin( 1x

)− cos

( 1x

).

Since limx→0

f ′(x) does not exist, then f ′(x) is not continuous at x = 0.

The Lagrange Mean Value Theorem is one of the most important result in calculus. The following

propositions are some Corollaries from this theorem.

Corollary 4.4.4 Suppose that f(x) is differentiable in (a, b). Then f(x) is constant if and only if f ′(x) = 0

in whole (a, b).

Proof. f(x) = C then obviously f ′(x) = 0 in (a, b). To prove the converse, let u and v be any two points

in (a, b) with u < v. Since f(x) is continuous on [u, v] and differentiable in (u, v). According to the Mean

Value Theorem, there is a point x0 ∈ (u, v) such that

f ′(x0) =f(v)− f(u)

v − u.

But f ′(x0) = 0, and thus f(u) = f(v). Consequently, f(x) is a constant.

From Corollary 1, we get the following corollary immediately.

Corollary 4.4.5 Suppose that f(x) and g(x) are differentiable in (a, b). Then they differ by a constant, i.e.,

f(x) = g(x) + C, if and only if g ′(x) = f ′(x).

Corollary 4.4.6 Suppose that f(x) is differentiable and f ′(x) > 0 on the interval (a, b). Then f(x) is

strictly increasing on (a, b). Moreover, if f(x) is continuous on [a, b], then f(x) is strictly increasing on

[a, b].

Proof. Let u and v be any two points with u < v in (a, b). Since f(x) is continuous on [u, v] and

differentiable in (u, v), we apply the Mean Value Theorem to choose a point x 0 ∈ (u, v) at which

f ′(x0) =f(v)− f(u)

v − u.

Since f ′(x0) > 0 and v − u > 0, it follows that f(v) > f(u).

Replacing the assumption f ′(x) > 0 by f ′(x) < 0, the conclusion is f(x) strictly decreasing in (a, b).

Theorem 4.4.7 (Cauchy Mean Value Theorem) Suppose that the functions f(x) and g(x) are continuous

on [a, b] and differentiable in (a, b). Moreover, assume that g ′(x) �= 0 ∀x ∈ (a, b). Then there is a point

x0 ∈ (a, b) at which

f(b)− f(a)

g(b)− g(a)=f ′(x0)g′(x0)

.


Proof. Since g ′(x) �= 0 in whole (a, b), it follows from the Lagrange Mean Value Theorem that g(b) −g(a) �= 0. Now define an auxiliary function ψ on [a, b] by

ψ(x) = f(x)− f(a)−[f(b)− f(a)

g(b)− g(a)

][g(x)− g(a)].

ψ(x) is continuous on [a, b] and differentiable in (a, b). Furthermore,

ψ(a) = ψ(b) = 0.

Therefore, we can apply Rolle’s Theorem to choose a point x 0 ∈ (a, b) at which ψ′(x0) = 0, namely

f ′(x0)− f(b)− f(a)

b− ag′(x0) = 0 =⇒ f(b)− f(a)

b− a=f ′(x0)g′(x0)

.

4.5 High (Order) Derivatives and Differential

Definition 4.5.1 Let f ′(x) be the derivative of f(x) in (a, b) and also differentiable in (a, b). Then,

the derivative of f ′(x) is called the second (two) derivative of f(x), denoted by f ′′(x) or f (2)(x), i.e.,

(f ′(x))′ = f ′′(x). Iteratively, if f (k)(x) is differentiable in (a, b) then the derivative of f (k)(x) is called

the (k + 1) (or (k + 1)th) derivative of f(x) denoted by f (k+1)(x), i.e., [f (k)(x)]′ = f (k+1)(x).

Theorem 4.5.2 (Leibniz’s Formula) Let I be an open interval and n be a natural number. Suppose that

both f : I → R and g : I → R have n derivatives, then fg : I → R has n derivatives, and

(fg)(n)(x) =

n∑k=0

( n

k

)f (k)(x)g(n−k)(x) ∀x ∈ I.

Notation of Leibnitz. An alternate notation, due to Leibnitz, for derivative f ′(x) is as following

f ′(x) =df

dxor

d

dx(f(x))

ordy

dx= y′(x) if y = y(x) or f(x)

Advantages. For numbers a, b we have ab = 1/ b

a . If y = f(x) has its inverse x = f−1(y) and f(x) is

differentiable, then according to the related theorem we have

(f−1)′(y) =1

f ′(x)

Using the notation of Leibnitz, we can write

dx

dy= 1/

dy

dx


For composite function h(x) = f(u(x)), we have the Chain Rule

h′(x) = f ′(u)u′(x)

Now we can write

df

dx=

df

du

du

dx

that is just like for numbers a, b �= 0, c �= 0

a

b=a

c· cb

For second derivative or high derivative, we denote f ′′(x) or f (k)(x) by

f ′′(x) =d2f

dx2or f (k)(x) =

dkf

dxk.

Definition 4.5.3 Suppose that y = f(x) is defined in some neighborhood of x. Let Δy = f(x+Δx)−f(x).If

Δy = AΔx+ α(Δx),

where A is independent of Δx, and

limΔx→0

α(Δx)

Δx= 0,

then AΔx is called the differential of y, and denoted by

dy = AΔx.

Clearly, if f ′(x) exists, then A = f ′(x). The reverse is also true: if ∃A s.t. Δy = AΔx+α(Δx), then

f ′(x) = A. Since dx = Δx, then

dy = Adx = f ′(x)dx.

All the formulas for derivatives can be transferred to differentials, e.g.,

d(u ± v) = du± dv

d(uv) = vdu + udv

d(uv

)=vdu− udv

v2

df(g(x)) = f ′(g(x))g′(x)dx, or, df(u) = f ′(u)du = f ′(u)g′(x)dx


4.6 Applications of Derivative

4.6.1 L’Hopital’s Rule

Theorem 4.6.1 Suppose that f(x) and g(x) are differentiable in a neighborhood of a, except possibly at a,

g′(x) �= 0, and

limx→a

f ′(x)g′(x)

= A (or ±∞).

If limx→a

f(x) = limx→a

g(x) = 0, then

limx→a

f(x)

g(x)= A.

Proof. Define f(a) = g(a) = 0, then f(x) and g(x) are continuous in the neighborhood. If x < a, then

from Cauchy Mean Value Theorem there exists a ξ ∈ (x, a) such that

limx→a−0

f(x)

g(x)= lim

x→a−0

f(a)− f(x)

g(a)− g(x)= lim

x→a−0

f ′(ξ)g′(ξ)

.

Since x→ a− 0, then ξ → a− 0. Thus, we have

limx→a−0

f(x)

g(x)= A.

Similarly, we can prove

limx→a+0

f(x)

g(x)= A.

Theorem 4.6.1’ Suppose that f(x) and g(x) are differentiable in a neighborhood of a, except possibly at

a, g′(x) �= 0, and

limx→a

f ′(x)g′(x)

= A (or ±∞).

If limx→a

f(x) = +∞, and limx→a

g(x) = +∞, then

limx→a

f(x)

g(x)= A.

Proof. Since g ′(x) �= 0, we may assume that g ′(x) < 0. So g(x) is a decreasing function. Since limx→a

g(x) =

+∞, we may assume that g(x) > 0 in a small interval (a, b]. Given ε > 0, ∃δ1, s.t.,

∣∣∣f ′(x)g′(x)

−A∣∣∣ < ε

2, ∀a < x < a+ δ1 ≡ x0.

Using Cauchy Mean Value Theorem on [x, x0], we get

f(x)− f(x0)

g(x)− g(x0)=f ′(c)g′(c)

, x < c < x0.

Then we have ∣∣∣f(x)− f(x0)

g(x)− g(x0)− A∣∣∣ < ε

2, ∀a < x < x0.


Since

f(x)

g(x)−A =

f(x)− f(x0)

g(x)− g(x0)· g(x)− g(x0)

g(x)+f(x0)

g(x)−A

=[f(x)− f(x0)

g(x)− g(x0)−A]·[g(x)− g(x0)

g(x)

]+A[g(x)− g(x0)

g(x)

]−A+

f(x0)

g(x)

=[1− g(x0)

g(x)

]·[f(x) − f(x0)

g(x) − g(x0)−A]+f(x0)−Ag(x0)

g(x),

then we have ∣∣∣f(x)g(x)

−A∣∣∣ ≤ ε

2+∣∣∣f(x0)−Ag(x0)

g(x)

∣∣∣.Since lim

x→ag(x) = +∞, ∃δ2, s.t.,

∣∣∣f(x0)−Ag(x0)

g(x)

∣∣∣ < ε

2, ∀a < x < a+ δ2.

Therefore, ∣∣∣f(x)g(x)

−A∣∣∣ ≤ ε

2+ε

2= ε,

i.e.,

limx→a+

f(x)

g(x)= A.

Similarly,

limx→a−

f(x)

g(x)= A.

For the case x→ ∞, we have

Theorem 4.6.2 Suppose that f(x) and g(x) are differentiable for |x| > a > 0, g ′(x) �= 0, and

limx→∞

f ′(x)g′(x)

= A (or ±∞).

If limx→∞ f(x) = lim

x→∞ g(x) = 0, then

limx→∞

f(x)

g(x)= A.

Proof. Let F (t) = f(1/t) and G(t) = g(1/t). Then

limt→0

F ′(t)G′(t)

= limt→0

f ′(1/t)(−1/t2)

g′(1/t)(−1/t2)= A.

Therefore, from Theorem 4.6.1 we get

limx→∞

f(t)

g(t)= lim

t→0

F (t)

G(t)= A.

Theorem 4.6.2’ Suppose that f(x) and g(x) are differentiable for |x| > a > 0, g ′(x) �= 0, and

limx→∞

f ′(x)g′(x)

= A (or ±∞).


If limx→∞ f(x) = +∞, and lim

x→∞ g(x) = +∞, then

limx→∞

f(x)

g(x)= A.

Using L’Hopital’s Rule, we can find the following limits easily:

limx→0

sinx

x, lim

x→0

ln(1 + x)

x, lim

x→0

ex − 1

x.

Example 4.6.1 ( 00 ) limx→0

sinx

x= 1.

Example 4.6.2 (∞∞ ) limx→+∞

lnx

x2= 0.

Example 4.6.3 (0 · ∞) limx→0+

x ln x = 0.

Example 4.6.4 (∞−∞) limx→0

( 1

x2− cos2 x

sin2 x

)=

1

3.

Example 4.6.5 (1∞) limx→0

(sinxx

) 11−cos x

= e−13 .

Example 4.6.6 limx→+∞

x− sinx

x+ sinx= 1.

4.6.2 Extreme values of functions

By using high derivatives, we have the following important theorem related to local minimizer and

local maximizer.

Theorem 4.6.3 Suppose that function f(x) has second derivative in (a, b) and at the point x 0 ∈ (a, b),

f ′(x0) = 0 (stationary point). Then

if f ′′(x0) > 0, x0 is a local minimizer;

if f ′′(x0) < 0, x0 is a local maximizer.

Proof. First suppose that f ′′(x0) > 0. Since

f ′′(x0) = limx→x0

f ′(x) − f ′(x0)x− x0

> 0.

it follows that there is a δ > 0 such that

f ′(x)− f ′(x0)x− x0

> 0 if 0 < |x− x0| < δ.

But f ′(x0) = 0, hence

f ′(x) > 0 if x0 < x < x0 + δ,


and

f ′(x) < 0 if x0 − δ < x < x0.

From Corollary 4.4.6 we have

f(x) > f(x0) if 0 < |x− x0| < δ.

A similar argument applies when f ′′(x0) < 0.

Lemma 4.6.4 Suppose that function f has n derivatives in (a, b) and at the point x 0 ∈ (a, b)

f (k)(x0) = 0, k = 0, 1, · · · , n− 1.

Then for each point x �= x0 in (a, b), there is a point ξ strictly between x and x0 at which

f(x) =f (n)(ξ)

n!(x− x0)

n.

Proof. Define g(x) = (x− x0)n Then

g(k)(x0) = 0, k = 0, 1, · · · , n− 1.

Then f and g have n derivatives and satisfy

f (k)(x0) = g(k)(x0) = 0, k = 0, 1, · · · , n− 1.

From the proof of Theorem 4.6.1 (repeatedly) we see that there exists a ξ such that

f(x)

g(x)=f (n)(ξ)

g(n)(ξ),

ξ is between x and x0. Since g(n)(x) = n!, we obtain

f(x) =f (n)(ξ)

n!(x− x0)

n.

Theorem 4.6.5 Suppose that function f(x) has continuous n derivatives in (a, b) and at the point x 0 ∈(a, b),

f (k)(x0) = 0, k = 0, 1, · · · , n− 1, f (n)(x0) �= 0.

Then if n is odd, x0 is not a local maximizer or minimizer; if n is even, x0 is a local local maximizer or

minimizer, and

1. if f (n)(x0) > 0, x0 is a local minimizer;

2. if f (n)(x0) < 0, x0 is a local maximizer.


Proof. Let h(x) = f(x)− f(x0). Then

h(k)(x0) = 0, k = 0, 1, · · · , n− 1.

From the above theorem, we know that in interval x0 − δ < x < x0 + δ with (x0 − δ, x0 + δ) ⊂ (a, b)

h(x) =h(n)(ξ)

n!(x− x0)

n, x0 − δ < ξ < x0 + δ,

or,

f(x)− f(x0) =f (n)(ξ)

n!(x − x0)

n, x0 − δ < ξ < x0 + δ.

Because f (n)(x) is continuous at x = x0, we may choose δ such that f (n)(x) has a same sign with f (n)(x0)

in the interval x0 − δ < x < x0 + δ. Therefore, If n is odd, in the interval (x0 − δ, x0 + δ)

f(x)− f(x0) =f (n)(ξ)

n!(x− x0)

n ξ ∈ (x0 − δ, x0 + δ),

f (n)(ξ) has a same sign with f (n)(x0). f(x) − f(x0) will change its sign from x < x0 to x > x0. Thus

x0 is not a local local maximizer or minimizer. If n is even, due to (x−x0)n

n! ≥ 0, f(x) − f(x0) has a same

sign with f (n)(x0) in (x0 − δ, x0 + δ).

1. If f (n)(x0) > 0, f(x) − f(x0) =f(n)(ξ)

n! (x − x0)n ≥ 0 in (x0 − δ, x0 + δ), hence x0 is a local

minimizer;

2. If f (n)(x0) < 0, f(x) − f(x0) =f(n)(ξ)

n! (x − x0)n ≤ 0 in (x0 − δ, x0 + δ), hence x0 is a local

maximizer.

4.6.3 Concavity and graph of functions

Lemma 4.6.6 Suppose that the function f(x) has continuous second derivatives on (a, b). Then ∀x 0, x1 ∈(a, b), ∃η, between x0 and x1, s.t.

f(x1) = f(x0) + (x1 − x0)f′(x0) +

f ′′(η)2

(x1 − x0)2.

Proof. Let

ψ(x) = f(x)− f(x0)− (x− x0)f′(x0),

φ(x) =1

2(x− x0)

2.

Then, using Cauchy Mean Value Theorem we get

f(x1)− f(x0)− (x1 − x0)f′(x0)

12 (x1 − x0)2

=f ′(ξ)− f ′(x0)

ξ − x0= f ′′(η),

where η is between x0 and x1.


Lemma 4.6.7 Suppose that the function f(x) has continuous second derivatives on (a, b), and f ′′(x) ≥0, ∀x ∈ (a, b). Then for any x0 ∈ (a, b), the tangent line of f(x) at (x0, f(x0)) is below the graph of f(x).

Proof. The tangent line at (x0, f(x0)) is

y = f(x0) + (x− x0)f′(x0).

Let (x1, y1) be any point on the tangent line and (x1, f(x1)) be the corresponding point on f(x). Then

from Lemma 4.6.6, we get

f(x1) = f(x0) + (x1 − x0)f′(x0) +

f ′′(η)2

(x1 − x0)2

≥ f(x0) + (x1 − x0)f′(x0) = y1.

Definition 4.6.8 Suppose that the function f(x) is defined on I . If ∀x 0, x1 ∈ I, x0 < x1, and ∀t ∈ (0, 1)

f [(1− t)x0 + tx1] ≤ (1− t)f(x0) + tf(x1),

then f(x) is called concave up on (a, b). If ∀x0, x1 ∈ I, x0 < x1, and ∀t ∈ (0, 1)

f [(1− t)x0 + tx1] ≥ (1− t)f(x0) + tf(x1),

then f(x) is called concave down on (a, b).

Theorem 4.6.9 Suppose that function f(x) has continuous second derivatives on (a, b) and f ′′(x) ≥0, ∀x ∈ (a, b). Then f(x) is concave up on (a, b). If f ′′(x) ≤ 0, ∀x ∈ (a, b). Then f(x) is concave

down on (a, b).

Proof. ∀x ∈ (a, b), from Lemma 4.6.7, we have

f(x) ≥ f(u) + (x− u)f ′(u), ∀u ∈ (a, b).

Then ∀x0, x1 ∈ (a, b), we get

f(x0) ≥ f(xt) + (x0 − xt)f′(xt),

f(x1) ≥ f(xt) + (x1 − xt)f′(xt).

Multiplying the first inequality by (1− t), the second inequality by t, and letting x t = (1− t)x0 + tx1, we

get the result.

Example 4.6.7 Sketch the graph of f(x) =1

32(3x5 − 20x3).

4.7 Exercise

1. Let f(x) = x3 + 2x+ 1. Find the equation of the tangent line to the graph of f at the point (2, 13).


2. Let f(x) =

⎧⎨⎩ m1x+ 4, x ≤ 0,

m2x+ 4, x > 0,where m1 �= m2 are constants. Show that f : R → R is

continuous but not differentiable at x = 0.

3. Use the definition of derivative to compute the derivative of the following functions at x = 1:

(a) f(x) =√x+ 1.

(b) f(x) = x3 + 2x.

(c) f(x) =1

1 + x2.

4. Let I and J be open intervals, and the function f : I → R and h : J → R have the property that

h(J) ⊆ I , so the composition f ◦ h : J → R is defined. Show that if x0 is in J , h : J → R is

continuous at x0, h(x) �= h(x0) if x �= x0, and f : I → R is differentiable at h(x0), then

limx→x0

f(h(x)) − f(h(x0))

h(x) − h(x0)= f ′(h(x0)).

5. Let f(x) =

⎧⎨⎩ 0, x ≤ 0,

xn, x > 0,Prove that f : R → R is differentiable for n ≥ 2.

6. Suppose that the function f : R → R has the property that −x2 ≤ f(x) ≤ x2, ∀x. Prove that f is

differentiable at x = 0 and that f ′(0) = 0.

7. Let g(x) =

⎧⎨⎩ 3x2, x ≤ 1,

a+ bx, x > 1,For what values of a and b is the function g : R → R differentiable

at x = 1.

8. Suppose that the function g : R → R is differentiable at x = 0. Also, suppose that for each natural

number n, g(1/n) = 0. Prove that g(0) = 0 and g ′(0) = 0.

9. Suppose that the function f : R → R is differentiable and monotonically increasing. Show that

f ′(x) ≥ 0 for all x.

10. Suppose that the function f : R → R is differentiable and that there is a bounded sequence {x n}with xn �= xm, if n �= m, such that f(xn) = 0 for every index n. Show that there is a point x0 at

which f(x0) = 0 and f ′(x0) = 0. Hint. Use the Sequential Compactness Theorem.

11. Suppose that the function f : R → R is differentiable at x0.

Prove that limx→x0

xf(x0)− x0f(x)

x− x0= f(x0)− x0f

′(x0).

12. Suppose that the function f : R → R is differentiable at x = 0. Prove that limx→0

f(x2)− f(0)

x= 0.

13. Suppose that the function f : R → R is differentiable at x = 0. For real numbers a, b, and c, with

c �= 0, Prove that limx→0

f(ax)− f(bx)

cx=[a− b

c

]f ′(0).


14. Let the functionh : R → R be bounded. Define the function f : R → R by f(x) = 1+4x+x 2h(x).

Prove that f(0) = 1 and f ′(0) = 4. Note: There is no assumption about the differentiability of the

function h.

15. For a natural number n, the Geometric Sum Formula asserts that

1 + x+ · · ·+ xn =1− xn+1

1− x, x �= 1.

By differentiating, find a formula for 1+x+2x2+ · · ·+nxn and then for 12+22x+ · · ·+n2xn−1.

16. Let f(x) = x2 sin1

x, x �= 0, and f(0) = 0. Show that f ′(x) is not continuous.

17. Let f(x) = xn sin1

x, x �= 0, and f(0) = 0. Under what conditions, (1) f(x) is continuous at x = 0;

(2) f ′(x) exists at x = 0; (3) f ′(x) is continuous at x = 0.

18. Let f(x) = |x|n sin1

|x|m , x �= 0, and f(0) = 0 (m > 0). Under what conditions, (1) f ′(x) is

bounded in some neighborhood of x = 0; (2) f ′(x) is not bounded in any neighborhood of x = 0.

19. Let f(x) = (x− a)ϕ(x), where ϕ(x) is continuous at x = a, Find f ′(a).

20. Let f(x) = |x − a|ϕ(x), where ϕ(x) is a continuous function and at ϕ(a) �= 0. Show that f ′(a)

does not exist. What are f ′−(a) and f ′

+(a).

21. Let f(x) = x2∣∣∣ cos π

x

∣∣∣, x �= 0, and f(0) = 0. Show that f ′(0) exists, but in any neighborhood of 0,

there exists a point x0, such that f ′(x0) does not exist.

22. Let f(x) = x2, if x is rational, and f(x) = 0, if x is irrational, Show that f(x) is differentiable only

at x = 0.

23. Discuss whether the following functions are differentiable:

(a) y = |(x− 1)(x− 2)2(x− 3)3|.

(b) y = | cosx|.

(c) y = |π2 − x2| sin2 x.

(d) y = sin−1(cos x).

(e) y =

⎧⎨⎩

x−14 (x+ 1)2, |x| ≤ 1,

|x| − 1, |x| > 1,

24. Find the left and right derivative f ′−(x) and f ′

+(x):

(a) f(x) = |x|.

(b) f(x) = [x] sinπx.

(c) f(x) = x∣∣∣ cos π

x

∣∣∣.(d) f(x) = x

1+e1x, x �= 0; f(0) = 0.

(e) f(x) =√1− e−x2 .

(f) f(x) = | ln |x||, x �= 0.

(g) f(x) = sin−1 2x1+x2 .

(h) f(x) = (x− 2) tan−1 1x−2 , x �= 2; f(2) = 0.

25. Let f(x) = x sin1

x, x �= 0, and f(0) = 0. Show that f(x) is continuous at x = 0, but f(x) has no

left and right derivatives.


26. Let f(x) = x2, x ≤ x0, and f(x) = ax + b, x > x0. For what a and b f(x) is continuous and

differentiable at x = x0?

27. Let F (x) = f(x), x ≤ x0, and f(x) = ax+ b, x > x0, where f(x) has left derivative. For what a

and b F (x) is continuous and differentiable at x = x0?

28. Find the formula for the sum:

Pn = 1 + 2x+ 3x2 + · · ·+ nxn−1, Qn = 12 + 22x+ 32x2 + · · ·+ n2xn−1.

29. Find the formula for the sum:

Sn = sinx+ sin 2x+ · · ·+ sinnx, Tn = cosx+ 2 cos 2x+ · · ·+ n cosnx.

30. Use the identity cosx

2cos

x

4· · · cos x

2n=

sinx

2n sin x2n

, to find the formula for the sum:

1

2tan

x

2+

1

4tan

x

4+ · · ·+ 1

2ntan

x

2n.

31. Show that the derivative of an even function is an odd function, and the derivative of an odd function

is an even function.

32. Show that the derivative of a periodic function is still a periodic function with the same period.

33. Find the derivative:

(a)(1− x)p

(1− x)q, x �= −1. (b) 2tan(1/x).

34. Let f(x) =1√

1 + x2∀x > 0. Find (f−1)′(

√1/5).

35. Let I be a neighborhood of x0 and let f : I → R be continuous, strictly monotone, and differentiable

at x0. Assume that f ′(x0) = 0. Use the characteristic property of inverse, f −1(f(x)) = x, ∀x, and

the Chain Rule to prove that the inverse function f −1 : f(I) → R is not differentiable at f(x0).

36. Suppose that the function f : R → R is differentiable and that {xn} is a strictly increasing bounded

sequence with f(xn) ≤ f(xn+1) ∀n ∈ N. Prove that there is a number x0 at which f ′(x0) ≥ 0.

Hint. Apply the Monotone Convergence Theorem.

37. For real number a, b, c, and d, define Q = {x | cx+ d �= 0}. Then define f(x) =ax+ b

cx+ d, ∀x ∈ Q.

Show that if the function f : Q → R is not contant, then it fails to have any local maximizer or

minimizer. Sketch the graph.

38. For c > 0, prove that the following equation does not have two solutions: x 3−3x+c = 0, 0 < x < 1.

39. Prove that the following equation has exactly one solution: x 5 + 5x+ 1 = 0, −1 < x < 0.

40. Prove that the following equation has exactly two solutions: x 4 + 2x2 − 6x+ 2 = 0, x ∈ R.


41. Suppose that the function f : [a, b] → R is continuous and that the restriction of f to the open

interval (a, b) is differentiable. Also suppose that f(a) > a, f(b) < b, and f ′(x) �= 1 for all

x ∈ (a, b). Prove that there exists a unique point ξ ∈ (a, b) such that f(ξ) = ξ.

42. Suppose that the function f : (0, 1) → R is differentiable, but unbounded. Prove that f ′(x) is also

unbounded on (0, 1). Hint. If not, |f(x)| ≤ |f(x0)|+ |f ′(ξ)| · |x− x0|.

43. For any numbers a and b and an even natural number n, show that the following equation has at most

two solutions: xn + ax+ b = 0, x ∈ R. Is this true if n is odd?

44. For numbers a and b, prove that the following equation has exactly three solutions if and only if

4a3 + 27b2 < 0: x3 + ax+ b = 0, x ∈ R.

45. Let n be a number. Suppose that the function f : R → R is differentiable and that the equation

f ′(x) = 0, x ∈ R, has at most n − 1 solutions. Prove that the equation f(x) = 0, x ∈ R, has at

most n solutions.

46. Use an induction argument to prove that if p : R → R is a polynomial of degree n, then the equation

p(x) = 0, x ∈ R, has at most n solutions.

Hint: use the result of the previous problem.

47. Let g : R → R and f : R → R be differentiable functions and suppose that g(x)f ′(x) =

f(x)g′(x), ∀x. If g(x) �= 0, ∀x ∈ R, show that there is some c ∈ R such that f(x) = cg(x), ∀x ∈ R.

48. Define f(x) =

⎧⎨⎩ x− x2, if x is rational,

x+ x2, if x is irrational.Show that f ′(0) = 1 and yet there is no neighbor-

hood I of the point 0 on which this function is monotonically increasing.

49. Let the function f : R → R have the property that there is a positive number c such that |f(u) −f(v)| ≤ c(u− v)2, ∀u, v ∈ R. Prove that f is constant.

50. Suppose that the function f(x) is differentiable on (a,+∞), and limx→+∞ f ′(x) = 0. Prove that

limx→+∞

f(x)

x= 0.

Hint. Using mean value theorem we get∣∣∣f(x)

x− f(x0)

x

∣∣∣ =∣∣∣ f(x)− f(x0)

x

∣∣∣ =∣∣∣f ′(ξ)(x− x0)

x

∣∣∣. For given

ε > 0, since limx→+∞ f ′(x) = 0, ∃M > 0, s.t., |f ′(x)| < ε/2, ∀x > M . Take x0 > M , then if x > x0, we

have∣∣∣ f(x)

x− f(x0)

x

∣∣∣ < ε

2

∣∣∣x− x0

x

∣∣∣ < ε

2. Therefore,

∣∣∣ f(x)x

∣∣∣ ≤∣∣∣ f(x)

x− f(x0)

x

∣∣∣+∣∣∣f(x0)

x

∣∣∣ < ε

2+∣∣∣f(x0)

x

∣∣∣.Since lim

x→+∞f(x0)

x= 0, ∃M1, s.t.,

∣∣∣f(x0)

x

∣∣∣ < ε

2, ∀x > M1. Therefore,

∣∣∣ f(x)x

∣∣∣ ≤ ε

2+

ε

2= ε, ∀x >

max{x0,M1}, i.e., limx→+∞

f(x)

x= 0.

51. Suppose that f : R → R is differentiable and that there is a positive number c such that f ′(x) ≥c, ∀x. Prove that f(x) ≥ f(0) + cx if x ≥ 0 and f(x) ≤ f(0) + cx if x ≤ 0. Use these inequalities

to prove that f(R) = R.


52. Let the function f : R → R have two derivatives and suppose that f(x) ≤ 0 and f ′′(x) ≥ 0 for all

x. Prove that f is constant. Hint: Observe that f ′ is increasing.

53. Suppose that the function f : R → R has two derivatives, with f(0) = f ′(0) = 0 and |f ′′(x)| ≤ 1

if |x| ≤ 1. Prove that f(x) ≤ 1/2 if |x| ≤ 1.

54. Let p : R → R be a polynomial of degree no greater than 5. Suppose that at some point x 0 ∈ R,

p(x0) = p′(x0) = · · · = p(5)(x0) = 0. Prove that p(x) = 0 ∀x ∈ R.

55. Suppose that the function f : [a, b] → R and g : [a, b] → R are continuous and that their restrictions

to the open interval (a, b) are differentiable. Also suppose that |f ′(x)| ≥ |g′(x)| > 0 ∀x ∈ (a, b).

Prove that |f(u)− f(v)| ≥ |g(u)− g(v)| ∀u, v ∈ [a, b].

56. Suppose that the function f : (−1, 1) → R has n derivatives and that its nth derivative f (n) :

(−1, 1) → R is bounded. Assume also that f(0) = f ′(0) = · · · = f (n−1)(0) = 0. Prove that there

is a positive numberM such that |f(x) ≤M |x|n ∀x ∈ (−1, 1).

57. Suppose that the function f : (−1, 1) → R has n derivatives. Assume that there is a positive number

M such that |f(x)| ≤M |x|n ∀x ∈ (−1, 1). Prove that f(0) = f ′(0) = · · · = f (n−1)(0) = 0.

58. Let I be a neighborhood of x0 and suppose that the function f : I → R has two continuous

derivatives. Prove that

limh→0

f(x0 + h)− 2f(x0) + f(x0 − h)

h2= f ′′(x0).

59. Suppose that f(x) is differentiable in the finite or infinite interval (a, b), and

limx→a+

f(x) = limx→b−

f(x). Show that there exists a c ∈ (a, b) such that f ′(c) = 0.

60. Suppose that the function f(x) is continuous on [a, b] and differentiable on (a, b). If f(a) = 0 and

there exists a constant C > 0 such that |f ′(x)| ≤ C|f(x)|, ∀x ∈ (a, b). Show that f(x) ≡ 0 on

[a, b]. Hint. Let x0 = inf{x | f(x) �= 0, x ∈ [a, b], then it is easy to see that f(x0) = 0. If x0 �= b, then let x1 > x0,

and M = maxx∈[x0,x1]

|f(x)|. Assume that f(x) = M > 0, x ∈ [x0, x1]. Thenf(x)− f(x0)

x− x0= f ′(ξ), or

M = |f(x)| = |f ′(ξ)(x − x0)| ≤ C|f(ξ)|(x− x0) ≤ CM(x− x0), i.e., C(x− x0) ≥ 1, or x− x0 ≥ 1C

Take x1 such that x− x0 ≤ x1 − x0 < 1C

, we get a contradiction.

61. Suppose that f (n−1)(x) is continuous in the closed interval [x0, xn], f (n)(x) exists in the open

interval (x0, xn), and f(x0) = f(x1) = · · · = f(xn), x0 < x1 < · · · < xn. Show that there exists

a c ∈ (x0, xn) such that f (n)(c) = 0.

62. Suppose that f (p+q)(x) is continuous in the closed interval [a, b], f (p+q+1)(x) exists in the open

interval (a, b), and f(a) = f ′(a) = · · · = f (p)(a) = 0, f(b) = f ′(b) = · · · = f (q)(b) = 0. Show

that there exists a c ∈ (a, b) such that f (p+q+1)(c) = 0.


63. Suppose that Pn(x) = a0xn + a1x

n−1 + · · · + an, a0 �= 0, where ak, k = 0, 1, · · · , n are real

numbers, has only real roots. Show that P ′n(x), P

′′n (x), · · · , P (n−1)

n (x) also have only real roots.

64. Prove that the following equation has exactly two solutions:

x4 − 4x3 + 8x2 − 14x+ 11 = 0, 0 ≤ x < +∞.

65. Prove that if x ≥ 0, then√x+ 1−√

x =1

2√x+ θ(x)

, where1

4≤ θ(x) ≤ 1

2, and lim

x→0θ(x) =

1

4,

limx→+∞ θ(x) =

1

2.

66. Prove the inequalities:

(a) | sinx− sin y| ≤ |x− y|. (b)a− b

a< ln

a

b<a− b

b, 0 < b < a.

67. Prove the inequalities:

(a) x− x2/2 < ln(1 + x) < x, x > 0. (b) x− x3/6 < sinx < x, x > 0.

68. Suppose that f(x) is differentiable in the interval (x0,+∞) and limx→+∞ f ′(x) = 0.

Prove that limx→+∞

f(x)

x= 0.


(a) limx→0

tanx− x

x− sinx. (b) lim

x→+∞xn

eax, a > 0, n > 0.

70. Let f(x) = (x − a)nϕ(x), where the function ϕ(x) has continuous n − 1 derivatives in the neigh-

borhood of a. Find f (n)(a).

71. Find the local maximizer , local minimizer or stationary points of the following functions.

(a) x3 − x+ 5. (b) x9 − x7 + 11.

72. If (a) f(x) is differentiable at x0, but g(x) is not differentiable at x0; (b) both f(x) and g(x) are not

differentiable at x0, can you conclude that F (x) = f(x) + g(x) is not differentiable at x0?

73. If (a) f(x) is differentiable at x0, but g(x) is not differentiable at x0; (b) both f(x) and g(x) are not

differentiable at x0, can you conclude that F (x) = f(x)g(x) is not differentiable at x0? Consider

the example (a) f(x) = x, g(x) = |x|; (b) f(x) = |x|, g(x) = |x|.

74. If (a) f(x) is differentiable at x = g(x0), but g(x) is not differentiable at x0; (b) f(x) is not differ-

entiable at x = g(x0), but g(x) is differentiable at x0; (c) f(x) is not differentiable at x = g(x0),

and g(x) is not differentiable at x0, can you conclude that F (x) = f(g(x)) is not differentiable

at x0? Consider the example (a) f(x) = x2, g(x) = |x|; (b) f(x) = |x|, g(x) = x2; (c)

f(x) = 2x+ |x|, g(x) = 23x− 1

3 |x|.

75. Can f(x) have (a) finite derivative; (b) infinite derivative, at its discontinuous points? Consider the

example f(x) = sgnx.


76. If f(x) is differentiable on the finite interval (a, b) and limx→a

f(x) = ∞. Is it true limx→a

f ′(x) = ∞?

Consider the example f(x) =1

x+ cos

1

x, as x→ 0.

77. If f(x) is differentiable on the finite interval (a, b) and limx→a

f ′(x) = ∞. Is it true limx→a

f(x) = ∞?

Consider the example f(x) = 3√x, as x→ 0.

78. If f(x) is differentiable on the interval (x0,+∞) and limx→+∞ f(x) exists. Is it true lim

x→+∞ f ′(x)

exists? Consider the example f(x) =sin(x2)

x.

79. If f(x) is bounded, differentiable on the interval (x 0,+∞), and limx→+∞ f ′(x) exists. Is it true

limx→+∞ f(x) exists? Consider the example f(x) = cos(lnx).

Chapter 5

Indefinite Integral

5.1 Definition and Some Basic Formulas

5.1.1 Definition

Definition 5.1.1 Let f(x) be a function defined on [a, b]. If F (x) is differential on (a, b) and satisfies

F ′(x) = f(x) x ∈ (a, b),

then F (x) is called an antiderivative of f(x) and the set of all antiderivatives of f(x) is called the indefi-

nite integral of f(x) and denoted by ∫f(x)dx.

If F (x) and F (x) are two antiderivatives of f(x), then

[F (x) − F (x)]′ = F ′(x)− F ′(x) = f(x)− f(x) = 0.

By Corollary 4.4.5,

F (x) − F (x) = C, F (x) = F (x) + C.

Hence if F (x) is an antiderivative of f(x), the indefinite integral is∫f(x)dx = F (x) + C

where C is an arbitrary constant. By the definition for any constants α, β∫[αf(x) + βg(x)]dx = α

∫f(x)dx + β

∫g(x)dx.

Existence of the antiderivative. The existence of the antiderivative will be discussed in the next chapter.

The antiderivative may not exist for some functions.

103


Example 5.1.1 Let

f(x) =

⎧⎨⎩ 1, x ≤ 0,

0, x > 0.

Then there is no function F (x) such that F ′(x) = f(x), x ∈ (−∞,+∞).

Proof. By contradiction. Suppose that there exists a F (x), s.t., F ′(x) = f(x), x ∈ (−∞,+∞). Then we

have F ′(0) = f(0) = 1. Let Δx > 0 and consider the interval [0,Δx]. F (x) is continuous on [0,Δx] and

differentiable on (0,Δx). Using Lagrange Mean Value Theorem we get

F (0 + Δx)− F (0)

Δx= F ′(ξ) = f(ξ), 0 < ξ < Δx.

Therefore,

F ′+(0) = lim

Δx→0+

F (0 + Δx)− F (0)

Δxlim

ξ→0+f(ξ) = 0,

which contradicts with F ′(0) = 1.

Some Basic Formulas.

1.∫xαdx =

1

α+ 1xd+1 + C, α �= 1.

2.∫

dx

x= ln |x|+ C.

3.∫exdx = ex + C.

4.∫

cosxdx = sinx+ C.∫sinxdx = − cosx+ C.

5∫

sec2 xdx = tanx+ C.∫csc2 xdx = − cotx+ C.

6.∫

dx√a2 − x2

= sin−1(xa

)+ C.

7.∫

dx

a2 + x2=

1

atan−1

(xa

).

8.∫axdx =

1

ln aax + C, (a > 0, a �= 1).

5.2 Integration by Substitution

Suppose that F ′(y) = f(y), y = ψ(x). By the Chain Rule of differentiating composite function

d

dxF (ψ(x)) = F ′(y)ψ′(x) = f(ψ(x))ψ′(x).

Hence ∫f(ψ(x))ψ′(x)dx = F (ψ(x)) + C,

or, ∫f(ψ(x))dψ(x) =

∫f(y)dy = F (y) + C = F (ψ(x)) + C.

We will gain advantage from the formula if the antiderivative of f(y) is known and the function g(x) that

we are going to find its indefinite integral can be expressed as

g(x) = f(y(x))y′(x).

Chapter 5: Indefinite Integrals 105

Example 5.2.1

∫tanxdx =

∫sinx

cosxdx =

∫ −d(cosx)

cosx

= − ln | cosx|+ C.

Example 5.2.2 Let y = x2, then

∫xdx

1 + x4=

1

2

∫dy

1 + y2=

1

2tan−1 y + C =

1

2tan−1 x2 + C

Example 5.2.3

∫sinmx sinnxdx =

1

2

∫[cos(m− n)x− cos(m+ n)x]dx

=1

2

[1

m− n

∫cos(m− n)xd(m − n)x− 1

m+ n

∫cos(m+ n)xd(m + n)x

]

=1

2

[1

m− nsin(m− n)x− 1

m+ nsin(m+ n)x

]+ C, m �= n.

Similarly, we can find the following integrals:

∫sinmx cosnxdx,

∫cosmx cosnxdx.

Example 5.2.4

∫dx

a2 − x2=

1

2a

[∫dx

a+ x+

∫dx

a− x

]

=1

2a[ln |a+ x| − ln |a− x|+ C]

=1

2aln

∣∣∣∣a+ x

a− x

∣∣∣∣+ C

Example 5.2.5

∫dx

sinx=

∫dx

2 sin x2 cos

x2

=

∫sec2 x

2

2 tan x2

dx

=

∫d(tan x

2 )

tan x2

= ln∣∣∣tan x

2

∣∣∣+ C


Example 5.2.6 Let x = a sin t, then t = sin−1 xa , we have

∫ √a2 − x2dx =

∫a cos t · a cos tdt

= a2∫

cos2 dt = a2∫

1

2(1 + cos 2t)dt

=a2

2

[t+

1

2sin 2t

]+ C

=a2

2[t+ sin t cos t] + C

=a2

2

[sin−1 x

a+x

a

√1− (

x

a

2)

]+ C

=a2

2sin−1 x

a+x

2

√a2 − x2 + C

Example 5.2.7∫

dx√x2 + a2

= ln |x+√x2 + a2|+ C

Example 5.2.8∫

dx√x2 − a2

= ln |x+√x2 − a2|+ C

5.3 Integration by Parts

From the differentiation rule

[f(x) · g(x)]′ = f ′(x)g(x) + f(x)g′(x)

we get the formula for indefinite integral

∫f(x)g′(x)dx =

∫[f(x)g(x)]′dx−

∫f ′(x)g(x)dx,

or, ∫f(x)dg(x) = f(x)g(x) −

∫g(x)df(x).

This formula is called integration by parts. For some complicated integrals, the integral on the right hand

side may be a simplified one.

Example 5.3.1 Let In =

∫xnexdx, then we have

In =

∫xndex = xnex −

∫exd(xn)

= xnex − n

∫xn−1exdx = xnex − nIn−1


Therefore,

In = xnex − nIn−1

= xnex − nxn−1ex + n(n− 1)In−2

= · · ·= ex[xn − nxn−1 + n(n− 1)xn−2 − · · ·+ (−1)nn!] + C

Example 5.3.2 ∫xn lnxdx =

∫lnxd

(xn+1

n+ 1

)

=xn+1

n+ 1lnx−

∫xn+1

n+ 1d(lnx)

=xn+1

n+ 1lnx− 1

n+ 1

∫xn+1 · 1

xdx

=xn+1

n+ 1lnx− 1

n+ 1

∫xndx

=xn+1

n+ 1lnx− xn+1

(n+ 1)2+ C

Example 5.3.3 ∫tan−1 xdx = x tan−1 x−

∫xd(tan−1 x)

= x tan−1 x−∫

x

1 + x2dx

= x tan−1 x− 1

2

∫d(x2)

1 + x2

= x tan−1 x− 1

2ln(1 + x2) + C

Example 5.3.4 (Example 5.2.6)∫ √a2 − x2dx = x ·

√a2 − x2 −

∫xd√a2 − x2

= x ·√a2 − x2 +

∫x2dx√a2 − x2

= x ·√a2 − x2 −

∫a2 − x2√a2 − x2

dx+ a2dx√a2 − x2

= x ·√a2 − x2 −

∫ √a2 − x2dx+ a2 sin−1 x

a+ C

=⇒∫ √

a2 − x2dx =x

2

√a2 − x2 +

a2

2sin−1 x

a+ C

In some case we can’t apply the method of integration by parts to obtain the result directly, but we can

obtain a recurrent formula and then by this formula to get the result that we want.


Example 5.3.5 Find∫

dx

(x2 + a2)4.

We consider In =

∫dx

(x2 + a2)n

In =x

(x2 + a2)n−∫xd

1

(x2 + a2)n

=x

(x2 + a2)n+ 2n

∫x2

(x2 + a2)n+1dx

=x

(x2 + a2)n+ 2n In − 2na2 In+1

=⇒ In+1 =2n− 1

2na2In +

x

2na2(x2 + a2)n

It’s known that

I1 =

∫dx

x2 + a2=

1

atan−1 x

a+ C

Hence by the recurrent formula we can get I2, I3 and then the required result I4

5.4 Integration of Rational Functions


x+ 3

(x+ 1)(x− 2)2dx.

Letx+ 3

(x+ 1)(x− 2)2=

a

x+ 1+

b

x− 2+

c

(x − 2)2.

Then we have

x+ 3 = a(x− 2)2 + b(x+ 1)(x− 2) + c(x+ 1).

Compare the coefficients we get a = 2/9, b = −2/9, c = 5/3. Therefore,∫x+ 3

(x + 1)(x− 2)2dx = a

∫dxx+ 1

+ b

∫dxx− 2

+ c

∫dx

(x− 2)2

=2

9ln |x+ 1| − 2

9ln |x− 2| − 5

3· 1

x− 2+ C


6x2 − 15x+ 22

(x+ 3)(x2 + 2)2dx.

Let6x2 − 15x+ 22

(x+ 3)(x2 + 2)2=

a

x+ 3+bx+ c

x2 + 2+

dx + e

(x2 + 2)2.

Then we have

6x2 − 15x+ 22 = a(x2 + 2)2 + (bx+ c)(x+ 3)(x2 + 2) + (dx+ e)(x+ 3).


Compare the coefficients we get a = 1, b = −1, c = 3, d = −5, e = 0. Therefore,∫6x2 − 15x+ 22

(x+ 3)(x2 + 2)2dx =

∫dxx+ 3

+

∫ −x+ 3

x2 + 2dx+

∫ −5x

(x2 + 2)2dx

= ln |x+ 3| − 1

2ln |x2 + 2|+ 3√

2tan−1

( x√2

)+

5

2(x2 + 2)+ C

In general, to decompose a rational function f(x) = p(x)/q(x) into partial fractions, we proceed as

follows:

1. If f(x) is improper, i.e., p(x) is of degree at least that of q(x), divide p(x) by q(x), obtaining

p(x)

q(x)= g(x) +

r(x)

q(x).

2. Factor q(x) into a product of linear and irreducible quadratic factors with real coefficients.

3. For each factor of the form (ax+ b)k, expect the decomposition to have the terms

A1

(ax+ b)+

A2

(ax+ b)2+ · · ·+ Ak

(ax+ b)k.

4. For each factor of the form (ax2 + bx+ c)m, expect the decomposition to have the terms

B1x+ C1

(ax2 + bx+ c)+

B2x+ C2

(ax2 + bx+ c)2+ · · ·+ Bmx+ Cm

(ax2 + bx+ c)m.

5. Set r(x)/q(x) equal to the sum of all the terms found in Step 3 and 4. The number of constants to

be determined should equal the degree of the denominator q(x).

6. Multiply both sides of the equation found in Step 5 by q(x) and solve for the unknown constants.

This can be done by either of the two methods: (1) Equate coefficients of like-degree terms or (2)

assign convenient values to the variable x.

The above decompositions are based on the following two theorems:

Theorem 5.4.1 If q(x) = (ax+ b)kq1(x), q1(−b/a) �= 0, then

r(x)

q(x)=

Ak

(ax+ b)k+

r1(x)

(ax+ b)k−1q1(x),

where Ak is a constant, and r1(x) is a polynomial.

Proof. Only need to select Ak and r1(x) s.t.

r(x) = Akq1(x) + (ax+ b)r1(x).

Let x = −b/a, then we get Ak. Since r(−b/a)− Akq1(−b/a) = 0, r(x) − Akq1(x) has a factor ax+ b.

Then we get r1(x).


Theorem 5.4.2 If q(x) = (ax2 + bx+ c)mq1(x), and q1(x) has no factor ax2 + bx+ c, then

r(x)

q(x)=

Bmx+ Cm

(ax2 + bx+ c)m+

r1(x)

(ax2 + bx+ c)m−1q1(x),

where Bm and Cm are constants, and r1(x) is a polynomial.

Proof. Only need to select Bm, Cm, and r1(x) s.t.

r(x)

ax2 + bx+ c− (Bmx+ Cm)

q1(x)

ax2 + bx+ c= r1(x).

Suppose that

r(x)

ax2 + bx+ c= g(x) +

αx+ β

ax2 + bx+ c,

q1(x)

ax2 + bx+ c= h(x) +

γx+ δ

ax2 + bx+ c.

Then we have

r1(x) = g(x) + (Bmx+ Cm)h(x) − Bmγ

a

+

(α−Bmδ − Cmγ + Bmγ

a b)x+ β − Cmδ +

Bmγa c

ax2 + bx+ c.

There is a unique solution Bm and Cm such that r1(x) is a polynomial. .

In fact, it’s not true that we can find the indefinite integral for every elementary function. It can be

proved that some function looks quite simple such as

sinx

xor e−x2

but∫

sin xx dx or

∫e−x2

dx can’t be expressed by some formula of elementary functions.

5.5 Exercise

1. Evaluate the following indefinite integrals using substitution:

(a)∫x 3√x+ πdx.

(b)∫

dx√x2 + 2x+ 5

.

(c)∫x2 + 3x√x+ 4

dx.

(d)∫

dx

x2√x2 − 1

.

2. Evaluate the following indefinite integrals using integration by parts:

(a)∫x cos2 x sinxdx.

(b)∫

lnx

x2dx.

(c)∫

tan−1(1x

)dx.

(d)∫x2 cosxdx.

(e)∫eax cosxdx.

(f)∫

cos6 x sin2 xdx.

Chapter 6: Integration 111

3. Evaluate the following indefinite integrals using partial fractions:

(a)∫

3x3 − 8x+ 13

(x+ 3)(x− 1)2dx. (b)

∫1

(x− 1)2(x2 + 2x+ 6)2dx.


Chapter 6

Integration

6.1 Darboux Sums and Definition of Integral

6.1.1 Darboux Sum

If n is a natural number and

a = x0 < x1 < · · · < xn = b

then P = {x0, x1, · · · , xn} is called a Partition of the interval [a, b], and xi is called a partition point of P ,

the interval [xi−1, xi] is called a subinterval of partition P .

Suppose that function f(x) is defined on [a, b] and bounded. For a partition P we define

mi = inf{f(x) |x ∈ [xi−1, xi]} i = 1, · · · , nMi = sup{f(x) |x ∈ [xi−1, xi]} i = 1, · · · , n

and then define

L(f, P ) =n∑

i=1

mi(xi − xi−1)

U(f, P ) =

n∑i=1

Mi(xi − xi−1)

We call U(f, P ) and L(f, P ) the Upper Darboux Sum and the Lower Darboux Sum for the function f

based on the partition P .

6.1.2 Refinement Lemma

Lemma 6.1.1 Suppose that function f(x) is bounded on [a, b] and satisfies

m ≤ f(x) ≤M, for all x ∈ [a, b]

113


Then for any partition P of [a, b],

m(b− a) ≤ L(f, P ) ≤ U(f, P ) ≤M(b− a).

Proof. Since

m ≤ infa≤x≤b

{f(x)} ≤ infxi−1≤x≤xi

{f(x)} = mi

and

Mi = supxi−1≤x≤xi

{f(x)} ≤ supa≤x≤b

{f(x)} ≤M

and obviously

mi ≤Mi

Hence

n∑i=1

m(xi − xi−1) ≤n∑

i=1

mi(xi − xi−1) ≤n∑

i=1

Mi(xi − xi−1) ≤n∑

i=1

M(xi − xi−1)

Namely,

m(b− a) ≤ L(f, P ) ≤ U(f, P ) ≤M(b− a).

Definition 6.1.2 Partition P ∗ of [a, b] is called a refinement of P if each partition point of P is also a

partition point of P ∗.

Lemma 6.1.3 (Refinement Lemma) Suppose that the function f(x) is bounded on [a, b]. Let P be a

partition of [a, b] and P ∗ be a refinement of P . Then

L(f, P ) ≤ L(f, P ∗) ≤ U(f, P ∗) ≤ U(f, P ).

Proof. Let Pi be the partition of [xi−1, xi] that is induced by P ∗ and [xi−1, xi] is a subinterval of P . By

the above lemma 1 in each subinterval [xi−1 xi],

mi(xi − xi−1) ≤ L(f, Pi) ≤ U(f, Pi) ≤Mi(xi − xi−1)

Hence, we have

n∑i=1


i=1

L(f, Pi) ≤n∑

i=1

U(f, Pi) ≤n∑

i=1

Mi(xi − xi−1)

Butn∑

i=1

L(f, Pi) = L(f, P ∗) andn∑

i=1

U(f, Pi) = U(f, P ∗). Therefore

L(f, P ) ≤ L(f, P ∗) ≤ U(f, P ∗) ≤ U(f, P ).


Lemma 6.1.4 Suppose that function f(x) is bounded on [a, b]. Then for any two partitions P 1 and P2 of

[a, b],

L(f, P1) ≤ U(f, P2)

Proof. Let P ∗ be a partitions formed by taking the union of partition points of P 1 and P2. Hence P ∗ is a

common refinement of P1 and P2. From the Refinement Lemma

L(f, P1) ≤ L(f, P ∗) ≤ U(f, P ∗) ≤ U(f, P2).

6.1.3 Definition of Integral

Let L be the collection of all lower Darboux sums and U be the collection of all the upper Darboux

sums. From the Lemma 6.1.4, we know that any upper Darboux sum U(f, P ) is an upper bound of L.

Hence by Completeness Axiom, the set L has the least upper bound (denoted by I ∗ or I∗(f)) and

I∗ = supL ≤ U(f, P )

Similarly, any lower Darboux sum L(f,P) is a lower bound of the set U . Hence U is bounded below and has

the greatest lower bound (denoted by I ∗ or I∗(f)) and satisfies

L(f, P ) ≤ inf U = I∗

for any partition P . Therefore,

I∗ ≤ I∗.

I∗ is called the lower integral of f on [a, b], and I ∗ is called the upper integral of f on [a, b].

Definition 6.1.5 Suppose that function f(x) is defined on [a, b] and bounded. Then f(x) is said to be

integrable provided that

I∗ = I∗ = I,

and the number I is called the integral (or definite integral) of the function f(x) on the interval [a, b],

denoted by ∫ b

a

f or∫ b

a

f(x)dx.

If I∗ �= I∗, then we say that f(x) is not integrable on [a, b]. This integral is also called Riemann integral.

If a ≥ b and f is integrable on [b, a], we define∫ b

a

f(x)dx = −∫ a

b

f(x)dx,

∫ a

a

f(x)dx = 0.

Note. From the definition, the integral is independent of the variable under the integral, i.e.,∫ b

a

f(x)dx =

∫ b

a

f(t)dt.


6.1.4 Interpretation in Physics and Geometry

• If v(t) is the speed function of a car, then∫ t1

t0

v(t)dt is the distance covered by the car from time t0

to t1.

• If ρ(x) is the line density of a rod with length l, then∫ l

0

ρ(x)dx is the mass of the rod.

•∫ b

a

f(x)dx is the area of the region between the curve y = f(x) and the x-axis

x

y

y=f(x)

6.2 Integrability

6.2.1 Integrability Criterion

Theorem 6.2.1 Suppose that function f(x) is bounded on [a, b]. Then f(x) is integrable if and only if for

any positive number ε, there exists a partition P ∗ of [a, b] such that

U(f, P ∗)− L(f, P ∗) < ε

Proof. Part of ”if”: If for any ε > 0, the inequality

U(f, P ∗)− L(f, P ∗) < ε

holds for some partition P ∗, then we can prove

supL = inf U

That means f(x) integrable. For partition P ∗, we have the inequalities

L(f, P ∗) ≤ supL ≤ inf U ≤ U(f, P ∗)

Therefore, we obtain

0 ≤ inf U − supL ≤ U(f, P ∗)− L(f, P ∗) < ε


Because inf U and supL are two constant and ε is any positive number, the inequalities

0 ≤ (inf U − supL) < ε implies

inf U − supL = 0

i.e., inf U = supLPart of ”only if”: If inf U = supL = I , then we can prove for any ε > 0, there exists a partition P ∗ such

that

U(f, P ∗)− L(f, P ∗) < ε

Because I is the greatest lower bound of U , that means for a given ε > 0, there is an element U(f, P 1) in Usuch that

U(f, P1)− I <ε

2

Similarly I is the least upper bound of L, there is an element L(f, P2) in L such that

I − L(f, P2) = I <ε

2

Let P ∗ be formed by taking the union of partition points of P 1 and P2. Then P ∗ is common refinement of

P1 and P2. By Refinement Lemma,

U(f, P ∗) ≤ U(f, P1)

L(f, P ∗) ≥ L(f, P2)

Therefore

U(f, P ∗)− L(f, P ∗) ≤ U(f, P1)− L(f, P2)

= [U(f, P1)− I] + [I − L(f, P2)]

<ε

2+ε

2= ε


Example 6.2.1 Function f(x) is piecewise constant, meaning that there are points x 0, x1, . . . , xn with

a = x0 < x1 < · · · < xn = b and

f(x) = ci for x ∈ (xi−1, xi)

and at points xi, f(x) may be any value. Prove that f is integrable.

Proof. Because the number of xi is finite, f(x) is bounded. Let m,M be two real numbers such that

m ≤ f(x) ≤M and let h = min1≤i≤n

(xi − xi−1). Take partition P as follows,

P = {x0, x0 + δ, x1 − δ, x1 + δ, · · ·xn−1 − δ, xn+1 + δ, xn − δ, xn}


with δ ≤ h

3. (Note that δ ≤ h

3assures xi+1 − δ > xi + δ). For the partition P

U(f, P )− L(f, P ) ≤ (M −m)δ + (n− 1)2δ + δ

= 2n(M −m)δ

Hence for any ε > 0, take δ = min

{h

3,

ε

1 + 2n(M −m)

}, then for the P with this δ,

U(f, P )− L(f, P ) < ε.

By the Integrability Criterion, the piecewise constant function is integrable.

6.2.2 Two Kinds of Integrable Functions

Theorem 6.2.2 Suppose that f(x) is monotone on [a, b]. Then f is integrable.

Proof. Let Pn = {x0, x1, . . . , xn} be a partition with evenly spaced points. That is xi − xi−1 =b− a

nfor

all i. Pn is called Regular Partition. Since

mi = infxi−1≤x≤xi

{f(x)} = f(xi−1)

Mi = supxi−1≤x≤xi

{f(x)} = f(xi)

U(f, Pn)− L(f, Pn) =

n∑i=1

(Mi −mi)(xi − xi−1)

=b− a

n

n∑i=1

(f(xi)− f(xi−1))

=b− a

n(f(b)− f(a))

Hence, we can choose a sufficiently large n such that

U(f, Pn)− L(f, Pn) < ε.

Theorem 6.2.3 Suppose that f(x) is continuous on [a, b]. Then f is integrable.

Proof. f(x) is continuous in [a, b], then it is uniformly continuous. Hence we may choose a number δ > 0

such that for any given ε > 0

|f(u)− f(v)| < ε

b− a, if |u− v| ≤ δ

Now we choose a regular partitionPn with n such thatb− a

n≤ δ. For this Pn, the length of any subinterval

[xi−1, xi] isb− a

n≤ δ. By Extreme Value Theorem of continuous function, there are u i, vi ∈ [xi−1 xi]

such that Mi = f(ui),mi = f(vi). Then we obtain

Mi −mi = f(ui)− f(vi) <ε

b− a


due to

|ui − vi| ≤ (xi − xi−1) ≤ δ

Hence

U(f, Pn)− L(f, Pn) =

n∑i=1


<ε

b− a

n∑i=1

(xi − xi−1) = ε.

Corollary 6.2.4 Suppose that function f(x) is bounded on [a, b] and continuous in (a, b). Then f is inte-

grable.

Proof. Suppose f(x) ≤ M, ∀x ∈ [a, b], and let δ > 0 be a small number. Since f(x) is continuous on

[a+ δ, b− δ], then f(x) is integrable on [a+ δ, b− δ]. For given ε > 0, there exists a partition P ∗ (Theorem

6.2.1), s.t.,

U(f, P ∗)− L(f, P ∗) <ε

2.

By adding two subintervals [a, a+ δ] and [b− δ, b] to the partition P ∗, we get a partition P of [a, b]. Letma

andMa be the inf and sup of f(x) on [a, a+ δ], and let mb andMb be the inf and sup of f(x) on [b− δ, b],

then we have

U(f, P )− L(f, P ) = (Ma −ma)δ + U(f, P ∗)− L(f, P ∗) + (Mb −mb)δ

≤ 2Mδ +ε

2+ 2Mδ = 4Mδ +

ε

2.

Choose δ such that 4Mδ <ε

2, then we have

U(f, P )− L(f, P ) < ε.

From Theorem 6.2.1, f(x) is integrable on [a, b].

Example 6.2.2 f(x) =

⎧⎨⎩

sinπ

x, if 0 < x ≤ 1

4, if x = 0is integrable.

Note. The f(x) in this example is not continuous at x = 0.

6.2.3 Convergence of Darboux Sums

Definition 6.2.5 For a partition P = {x0, . . . xn} of interval [a, b], we define the Gap of P , denoted by

||P ||, the length of the largest subinterval induced by P . That is

||P || = max1≤i≤n

[xi − xi−1]


Then ||P || < α means every subinterval of partition P has length less than α.

Lemma 6.2.6 Suppose that function f(x) is bounded on [a, b]. Then f is integrable if and only if for any

ε > 0, there is a positive δ such that

U(f, P )− L(f, P ) < ε, if ||P || < δ

Proof. By the Integrability Criterion, the proof of ”part if” has completed immediately. We now prove the

converse. Suppose that f(x) is integrable. According to the Integrability Criterion, for given ε > 0, we can

choose a partition P ∗ = {x∗0, . . . , x∗n} of [a, b] such that

U(f, P ∗)− L(f, P ∗) <ε

2

ChooseM > 0 such that

−M ≤ f(x) ≤M, x ∈ [a, b]

We are now going to obtain the following estimate

U(f, P )− L(f, P ) ≤ 2nM ||P ||+ [U(f, P ∗)− L(f, P ∗)]

for any partition P . Once this estimate is proven, we take δ = ε/4nM . Then

U(f, P )− L(f, P ) < ε, if ||P || < δ

Let P = {x0, . . . , xm}. Separate the set of indices {1, . . . ,m} into two subsets A and B. Define A to

be the set of indices i such that the subinterval (xi−1, xi) contains a partition point of P ∗. Define B to

be the set of indices i such that the subinterval (xi−1, xi) doesn’t contain any partition point of P ∗. Since

Mi −mi ≤ 2M, xi − xi−1 ≤ ||P || and there are fewer than n indices in A,

∑i∈A

(Mi −mi)(xi − xi−1) ≤ 2nM ||P ||

On the other hand, if index i ∈ B, then the subinterval [x i−1 xi] is contained in one of the subintervals

[x∗j−1 x∗j ] induced by P ∗. Because in such subintervals [xi−1 xi],

∑[xx−1 xi]⊆[x∗

j−1 x∗j ]

(Mi −mi)(xi − xi−1) ≤[

supx∗j−1≤x≤x∗

j

{f(x)} − infx∗j−1≤x≤x∗

j

{f(x)}](x∗j − x∗j−1)

Hence ∑i∈B

(Mi −mi)(xi − xi−1) ≤ U(f, P ∗)− L(f, P ∗)


Consequently

U(f, P )− L(f, P ) =∑i∈A

(Mi −mi)(xi − xi−1) +∑i∈B


≤ 2nM ||P ||+ [U(f, P ∗)− L(f, P ∗)].

Theorem 6.2.7 (Darboux Sum Convergence Criterion) Suppose that function f(x) is bounded on [a, b].

Then f is integrable and∫ b

a

f(x)dx = A if and only if

lim||Pn||→0

L(f, Pn) = lim||Pn||→0

U(f, Pn) = A

Proof. part of ”only if”: Let {Pn} be any sequence of partitions of [a, b] with limn→∞ ||Pn|| = 0. Because f

is integrable, according the Lemma 6.2.6, for any ε > 0, we can choose δ > 0 such that for any partition P

U(f, P )− L(f, P ) < ε, if ||P || < δ

Since limn→∞ ||Pn|| = 0, we can choose a natural numberN such that

||Pn|| < δ if n ≥ N

Thus

0 ≤ U(f, Pn)− L(f, Pn) < ε, if n ≥ N.

By the definition of integral

L(f, Pn) ≤∫ b

a

f(x)dx ≤ U(f, Pn)

Consequently,

|∫ b

a

f(x)dx − L(f, Pn)| < ε,

|∫ b

a

f(x)dx − U(f, Pn)| < ε, n ≥ N

That is

lim||Pn||→0


U(f, Pn) =

∫ b

a

f(x)dx

part of ”if”: Because

lim||Pn||→0


U(f, Pn) = A

for any ε > 0, there is a natural numberN such that

U(f, Pn)− L(f, Pn) < ε, if n ≥ N


By the Integrable Criterion, f is integrable and satisfies

L(f, Pn) ≤∫ b

a

f(x)dx ≤ U(f, Pn)

Hence, by the Squeezing Principle, ∫ b

a

f(x)dx = A.

6.2.4 Convergence of Riemann Sums

Definition 6.2.8 Suppose that function f(x) is defined on [a, b] and P = {x 0, . . . , xn} is a partition of

[a, b]. Let ci ∈ [xi−1, xi]. Then the sum

n∑i=1

f(ci)(xi − xi−1)

is called a Riemann sum for the function f(x) based on the partition P . Denote it by R(f, P ).

Note. Riemann sum R(f, P ) depends on f(x) and partition P , and also depends on the choice of the c i’s.

But the notationR(f, P ) does not exhibit this dependence.

From the definitions of Riemann sum and Darboux sums, for each partition P ,

L(f, P ) ≤ R(f, P ) ≤ U(f, P )

Lemma 6.2.9 Suppose that function f(x) is bounded on [a, b]. Then for each partition P of [a, b] and each

positive ε, there are Riemann sums R(f, P ) and R ′(f, P ) (with suitable ci’s) such that

0 ≤ U(f, P )−R(f, P ) < ε

and 0 ≤ R′(f, P )− L(f, P ) < ε

Proof. It’s known that the upper Darboux sum

U(f, P ) =∑

Mi(xi − xi−1) where Mi = supxi−1≤x≤xi

{f(x)}

By the meaning of least upper bound,

Mi − ε

(b − a)

is not an upper bound of the set {f(x) |x ∈ [xi−1 xi]}. Therefore, there is a point ci ∈ [xi−1, xi] such that

f(ci) > Mi − ε

(b− a)

Hencen∑

i=1

[Mi − ε

(b − a)

](xi − xi−1) <

n∑i=1

f(ci)(xi − xi−1),


and then U(f, P )− ε < R(f, P ), or 0 ≤ U(f, P )−R(f, P ) < ε. Similarly we choose c ′i ∈ [xi−1 xi] such

that

f(c′i) < mi +ε

(b− a)

and then obtain

0 ≤ R′(f, P )− L(f, P ) < ε

where

R′(f, P ) =n∑

i=1

f(c′i)(xi − xi−1).

Theorem 6.2.10 (Riemann Sum Convergence Criterion) Suppose that f(x) is bounded on [a, b] and

{Pn} is any sequence of partition of [a, b] with limn→∞ ||Pn|| = 0. Then f is integrable and

∫ b

a

f(x)dx = A,

if and only if

limn→∞R(f, Pn) = A.

Proof. part of ”only if”: Suppose that f is integrable and∫ b

a

f(x)dx = A. From the Theorem of Darboux

Sum Convergence Criterion, we have

lim||Pn||→0


U(f, Pn) = A

But it’s known that

L(f, Pn) ≤ R(f, Pn) ≤ U(f, Pn)

By Squeezing Principle,

lim||Pn||→0

R(f, Pn) = A

part of ”if”: By the preceding Lemma 6.2.9, for ε = 1/nwe choose Riemann sumsR(f, P n) andR′(f, Pn)

such that

0 ≤ U(f, Pn)−R(f, Pn) <1

n

0 ≤ R′(f, Pn)− L(f, Pn) <1

n

Consequently, we have

R′(f, Pn)− 1

n< L(f, Pn) ≤ U(f, Pn) < R(f, Pn) +

1

n

By the assumption

limn→∞R(f, Pn) = lim

n→∞R′(f, Pn) = A

Hence

limn→∞L(f, Pn) = lim

n→∞U(f, Pn) = A

According to the Darboux Sum Convergence Criterion, f(x) is integrable and∫ b

a

f(x)dx = A.


6.3 Linearity, Monotonicity and Additivity

6.3.1 Linearity of Integral

Theorem 6.3.1 Suppose f(x) and g(x) are integrable on [a, b], then

∫ b

a

[αf(x) + βg(x)]dx = α

∫ b

a

f(x)dx + β

∫ b

a

g(x)dx.

Proof. By using the theorem on convergence of Riemann sums (or Darboux sums), we can prove the

identity immediately.

6.3.2 Monotonicity of Integral

Theorem 6.3.2 Suppose that f and g are integrable on [a, b] and

f(x) ≤ g(x), x ∈ [a, b]

then ∫ b

a

f(x)dx ≤∫ b

a

g(x)dx

Proof. For any partition P , U(f, P ) ≤ U(g, P ). Choose a sequence {Pn} with limn→0

||Pn|| = 0. Then we

obtain

limn→0

U(f, Pn) ≤ limn→0

U(g, Pn), i.e.,∫ b

a

f(x)dx ≤∫ b

a

g(x)dx.

Corollary 6.3.3 Suppose that f is integrable on [a, b], then∣∣∣∣∣∫ b

a

f(x)dx

∣∣∣∣∣ ≤∫ b

a

|f(x)|dx

Proof. Since f is integrable on [a, b], then from Theorem 6.2.1, for given ε > 0 there exists a partition

P = {x0, . . . , xn} of [a, b] such that

n∑i=1

(Mi(f)−mi(f))(xi − xi−1) < ε,

where Mi(f) is the sup of f(x) on [xi−1, xi], and mi(f) is the inf of f(x) on [xi−1, xi]. If mi(f) and

Mi(f) have the same sign, then

Mi(|f |)−mi(|f |) =Mi(f)−mi(f).


If Mi(f) ≥ 0, mi(f) ≤ 0, and Mi(f) ≤ |mi(f)|, then we have

Mi(|f |)−mi(|f |) = −mi(f)−mi(|f |) ≤ −mi(f) ≤Mi(f)−mi(f).

If Mi(f) ≥ 0, mi(f) ≤ 0, and Mi(f) ≥ |mi(f)|, then we have

Mi(|f |)−mi(|f |) =Mi(f)−mi(|f |) ≤Mi(f) ≤Mi(f)−mi(f).

Therefore, we have

n∑i=1

(Mi(|f |)−mi(|f |))(xi − xi−1) ≤n∑

i=1

(Mi(f)−mi(f))(xi − xi−1) < ε,

i.e., |f(x)| is integrable on [a, b]. Since

−|f(x)| ≤ f(x) ≤ |f(x)|, ∀x ∈ [a, b],

we have

−∫ b

a

|f(x)|dx ≤∫ b

a

f(x)dx ≤∫ b

a

|f(x)|dx,

or, ∣∣∣∣∣∫ b

a

f(x)dx

∣∣∣∣∣ ≤∫ b

a

|f(x)|dx.

Example 6.3.1 Let f(x) be defined on [0, 1] and

f(x) =

⎧⎨⎩ 1, x is rational,

−1, x is irrational,

then |f(x)| is integrable, but f(x) is not integrable.

Example 6.3.2 Suppose that f(x) and g(x) are integrable on [a, b]. Prove that f 2(x), g2(x), and f(x)g(x)

are integrable on [a, b].

Proof. Since f(x) and g(x) are integrable on [a, b], then there exist M > 0 such that

|f(x)| ≤M, |g(x)| ≤M, ∀x ∈ [a, b],

and a partition P = {x0, . . . , xn} of [a, b] such that

n∑i=1

(Mi(f)−mi(f))(xi − xi−1) <ε

2M,

n∑i=1

(Mi(g)−mi(g))(xi − xi−1) <ε

2M,


where Mi(f) is the sup of f(x) on [xi−1, xi], and mi(f) is the inf of f(x) on [xi−1, xi]. Given x′, x′′ ∈[xi−1, xi], we have

|f(x′′)g(x′′)− f(x′)g(x′)| = |(f(x′′)− f(x′))g(x′′) + f(x′)(g(x′′)− g(x′))|≤ M |f(x′′)− f(x′)|+M |g(x′′)− g(x′)|≤ M(Mi(f)−mi(f) +Mi(g)−mi(g)).

For given integer k > 0, there exist x′′ and x′ such that

Mi(fg)− 1

k< f(x′′)g(x′′),

mi(fg) +1

k> f(x′)g(x′).

Then we have

Mi(fg)−mi(fg)− 2

k< f(x′′)g(x′′)− f(x′)g(x′)

≤ M(Mi(f)−mi(f) +Mi(g)−mi(g)).

Since k is arbitrary, we have

Mi(fg)−mi(fg) ≤M(Mi(f)−mi(f) +Mi(g)−mi(g)),

and

n∑i=1

(Mi(fg)−mi(fg))(xi − xi−1)

≤n∑

i=1

(Mi(f)−mi(f))(xi − xi−1) +n∑

i=1

(Mi(g)−mi(g))(xi − xi−1) < ε,

i.e., f(x)g(x) is integrable on [a, b], and f 2(x) and g2(x) are also integrable on [a, b].

6.3.3 Additivity over Intervals

Proposition 6.3.4 If f is integrable on [a, b], and interval [c, d] ⊆ [a, b], then f is integrable on [c, d].

Proof. Because f is integrable on [a, b], we can choose a partition P on [a, b] such that the point c and d are

partition points of P and

U(f, P )− L(f, P ) < ε

where ε is any given positive number. Let P ′ be the partition on [c, d] induced from P , i.e., the partition

points of P ′ are the same as that of P in interval [c, d]. Hence

U(f, P ′)− L(f, P ′) ≤ U(f, P )− L(f, P ) < ε


Thus, f is integrable on [c, d].

Theorem 6.3.5 If f is integrable on [a, b], then f is integrable on [a, c] and [c, b], and furthermore,∫ b

a

f(x)dx =

∫ c

a

f(x)dx +

∫ b

c

f(x)dx, a < c < b.

Conversely if f is integrable on [a, c] and [c, b], then f is integrable on [a, b] and∫ c

a

f(x)dx +

∫ b

c

f(x)dx =

∫ b

a

f(x)dx

where c may be anywhere.

Proof. From Proposition 6.3.4, f is integrable on [a, c] and [c, b]. Let P ′n and P ′′

n be the regular partitions

of [a, c] and [c, b], and let Pn = P ′n ∪ P ′′

n . Then, R(f, Pn) = R(f, P ′n) + R(f, P ′′

n ), and

lim||Pn||→0

R(f, Pn) = lim||Pn||→0

R(f, P ′n) + lim

||Pn||→0R(f, P ′′

n ),

or, ∫ b

a

f(x)dx =

∫ c

a

f(x)dx+

∫ b

c

f(x)dx a < c < b.

Conversely, let P ′n and P ′′

n be the regular partitions of [a, c] and [c, b], and let Pn = P ′n ∪ P ′′

n . Since f is

integrable on [a, c] and [c, b], from Riemann Sum Convergence Theorem we get

lim||Pn||→0

R(f, Pn) = lim||Pn||→0

R(f, P ′n) + lim

||Pn||→0R(f, P ′′

n )

=

∫ c

a

f(x)dx+

∫ b

c

f(x)dx, a < c < b,

i.e., f is integrable on [a, b].

6.4 Fundamental Theorem of Calculus

6.4.1 First Fundamental Theorem of Calculus

Theorem 6.4.1 (First Fundamental Theorem) Suppose that f(x) is integrable on [a, b], and also suppose

that F (x) is continuous on [a, b], differentiable on (a, b), and

F ′(x) = f(x), x ∈ (a, b).

Then ∫ b

a

f(x)dx = F (b)− F (a).

Proof. Firstly we prove that for any partition P of [a, b]

L(f, P ) ≤ F (b)− F (a) ≤ U(f, P )


Let P = {x0, x1, . . . xn}. We apply the Lagrange Mean Value Theorem to function F on every subinterval

[xi−1, xi],

F (xi)− F (xi−1) = F ′(ci)(xi − xi−1)

= f(ci)(xi − xi−1), xi−1 < ci < xi

Since

mi = infxi−1≤x≤xi

{f(x)} ≤ f(ci) ≤ supxi−1≤x≤xi

{f(x)} =Mi

we obtain

mi(xi − xi−1) ≤ F (xi)− F (xi−1) ≤Mi(xi − xi−1)

and thenn∑

i=1


i=1

(F (xi)− F (xi−1)) ≤n∑

i=1

Mi(xi − xi−1)

or

L(f, P ) ≤ F (b)− F (a) ≤ U(f, P ).

Therefore,

supL ≤ F (b)− F (a) ≤ inf U

Since f(x) is integrable, supL = inf U =

∫ b

a

f(x)dx. Hence,

∫ b

a

f(x)dx = F (b)− F (a).

Note. The First Fundamental Theorem of Calculus gives the relation betwween the definite integral and

indefinite integral (antiderivative): if ∫f(x)dx = F (x) + c,

then ∫ b

a

f(x)dx =[ ∫

f(x)dx]ba.

Let S(t) be the displacement function of a particle. Then the distance covered by the particle from time

t = 0 to t = T is S(T )− S(0). One the other hand, the speed function v(t) of the particle is

v(t) = S′(t)

By using v(t) to calculate the distance covered by the particle in the time period [0, T ] is∫ T

0

v(t)dt. Hence

we know ∫ T

0

S′(t)dt = S(T )− S(0)


Example 6.4.1 Evaluate the integral∫ π

4

π6

xdx

1 + x4.

Solution. It’s easy to find the antiderivative of f(x) =x

1 + x4. That is

F (x) =1

2tan−1(x2)

By the First Fundamental Theorem we get,∫ π4

π6

xdx

1 + x4= F (

π

4)− F (

π

6) =

1

2

[tan−1(

π

4)2 − tan−1(

π

6)2]

Example 6.4.2 Find the limit of

limn→∞

√1 +

√2 + · · ·+√

n

n32

Solution. It’s the Riemann Sum of the function f(x) =√x in interval [0, 1]. Hence the limit is

∫ 1

0

√xdx =[

2

3x

32

]10

=2

3.

6.4.2 Mean Value Theorem for Integral

Lemma 6.4.2 Suppose that the functions f(x) and g(x) are bounded and integrable on [a, b], then the

function f(x)g(x) is also integrable on [a, b].

Theorem 6.4.3 (Mean Value Theorem for Integral) Suppose that the function f(x) is continuous on

[a, b] and that the function g(x) is integrable and does not change the sign on [a, b]. Then there is a point

x0 ∈ [a, b] such that ∫ b

a

f(x)g(x)dx = f(x0)

∫ b

a

g(x)dx.

Proof. Without loss of generality assume that g(x) ≥ 0. Since f(x) is continuous on [a, b], f(x) has a

maximumM and a minimumm. Then we have

mg(x) ≤ f(x)g(x) ≤Mg(x).

Since f(x) and g(x) are integrable, then f(x)g(x) is also integrable. Using Theorem 6.3.2 we get

m

∫ b

a

g(x)dx ≤∫ b

a

f(x)g(x)dx ≤M

∫ b

a

g(x)dx.

If∫ ba g(x)dx = 0, then

∫ ba f(x)g(x)dx = 0 from the above inequality, and result follows. If

∫ ba g(x)dx > 0,

then

m ≤∫ baf(x)g(x)dx∫ ba g(x)dx

≤M.


Then from the Intermediate Value Theorem there is a point x 0 ∈ [a, b] such that

f(x0) =

∫ baf(x)g(x)dx∫ ba g(x)dx

,

which is the result of the theorem.

In the special case when g(x) ≡ 1 we get

∫ b

a

f(x)dx = f(x0)(b− a).

6.4.3 Second Fundamental Theorem of Calculus

Proposition 6.4.4 Suppose that the function f(x) is integrable on [a, b]. Define

F (x) =

∫ x

a

f(t)dt ∀x ∈ [a, b]

Then the function F (x) is continuous on [a, b].

Proof. Since f is integrable, it is bounded. Choose M > 0 such that

|f(x)| ≤M, for x ∈ [a, b]

For any point x0 ∈ [a, b]

F (x)− F (x0) =

∫ x

a

f(t)dt−∫ x0

a

f(t)dt =

∫ x

x0

f(t)dt.

Then we have,

|F (x)− F (x0)| =

∣∣∣∣∫ x

x0

f(t)dt

∣∣∣∣ ≤∣∣∣∣∫ x

x0

|f(t)|dt∣∣∣∣

≤∣∣∣∣∫ x

x0

Mdt

∣∣∣∣ =M |x− x0|.

Hence

limx→x0

{F (x)− F (x0)} = 0.

Theorem 6.4.5 (Second Fundamental Theorem) Suppose that the function f(x) is continuous on [a, b].

Thend

dx

∫ x

a

f(t)dt = f(x) ∀x ∈ (a, b).

Proof. Let

F (x) =

∫ x

a

f(t)dt ∀x ∈ (a, b).


Using Mean Value Theorem we have

F (x+Δx)− F (x) =

∫ x+Δx

x

f(t)dt = f(ξ) ·Δx ∀x ∈ (a, b),

where ξ is between x and x+Δx. Hence as Δx→ 0, ξ → x. Therefore

F ′(x) = limΔx→0

F (x +Δx)− F (x)

Δx= f(x).

Note. The Second Fundamental Theorem gives the existence of antiderivative (indefinite integral): A con-

tinuous function on [a, b] has an antiderivative. However, if f(x) is not continuous, F (x) may not be

differentiable on (a, b).

Example 6.4.3 Function f(x) define on [−1, 1] as following

f(x) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

−1, −1 ≤ x < 0,

0, x = 0,

1, 0 < x ≤ 1.

Then

F (x) =

∫ x

−1

f(x)dx =

⎧⎨⎩ −x− 1, −1 ≤ x < 0,

x− 1, 0 ≤ x ≤ 1,

or

F (x) = |x| − 1.

F is not differentiable at x = 0.

Corollary 6.4.6 Suppose that f(x) is continuous on (a, b) and ϕ(x) is differentiable on (c, d), and satisfies

ϕ((c, d)) ⊆ (a, b). Then for a fixed point x0 ∈ (a, b) we have

d

dx

∫ ϕ(x)

x0

f(t)dt = f(ϕ(x))ϕ′(x) for x ∈ (c, d)

Proof. Let

G(x) =

∫ ϕ(x)

x0

f(t)dt ∀x ∈ (c, d),

and let F (x) be the antiderivative of f(x). Then

G(x) =

∫ ϕ(x)

x0

f(t)dt = F (y)− F (x0),

where y = ϕ(x). By the Chain Rule of composite function,

G′(x) =dF

dy· dydx

= f(y)ϕ′(x) = f(ϕ(x))ϕ′(x).


6.4.4 Calculation of Integral

In Chapter 5, there are two methods for calculating the antiderivative of a function: Integration by

Substitution and Integration by Parts. Through the First Fundamental Theorem, we can use these methods

for calculating integrals.

Theorem 6.4.7 (Integration by Parts) Suppose that the functions h and g are continuous, and that f ′ and

g′ exist and integrable on [a, b]. Then

∫ b

a

h(x)g′(x)dx = h(x)g(x)|ba −∫ b

a

g(x)h′(x)dx.

Proof. Since

(h(x)g(x))′ = h(x)g′(x) + g(x)h′(x) ∀x ∈ (a, b),

hg′ and gh′ are integrable on [a, b], then by the First Fundamental Theorem we have

∫ b

a

[h(x)g′(x) + g(x)h′(x)]dx =

∫ b

a

[h(x)g(x)]′dx = h(x)g(x)|ba.

On the other hand,∫ b

a

[h(x)g′(x) + g(x)h′(x)]dx =

∫ b

a

h(x)g′(x)dx +

∫ b

a

g(x)h′(x)dx.

Combining the above two equations we get the result.

Example 6.4.4 (Second Mean Value Theorem for Integrals) Suppose that g(x) is continuous on [a, b],

f(x) is differentiable on [a, b]. Then we have the following results:

(a) If f ′(x) ≥ 0 and f(a) ≥ 0, then there exists a ξ ∈ [a, b], s.t.,

∫ b

a

f(x)g(x)dx = f(b)

∫ b

ξ

g(x)dx.

(b) If f ′(x) ≤ 0 and f(b) ≥ 0, then there exists a ξ ∈ [a, b], s.t.,

∫ b

a

f(x)g(x)dx = f(a)

∫ ξ

a

g(x)dx.

(c) In general, if f ′(x) ≥ 0 or f ′(x) ≤ 0, then there exists a ξ ∈ [a, b], s.t.,

∫ b

a

f(x)g(x)dx = f(a)

∫ ξ

a

g(x)dx + f(b)

∫ b

ξ

g(x)dx.

Proof. Case (a). Let

G(x) =

∫ b

x

g(t)dt,


thenG′(x) = −g(x), and∫ b

a

f(x)g(x)dx = −∫ b

a

f(x)dG(x)

= −f(x)G(x)|ba +

∫ b

a

G(x)f ′(x)dx

= f(a)G(a) +

∫ b

a

f ′(x)G(x)dx.

Let m andM be the minimum and maximum value of G(x) on [a, b], then

m(f(b)− f(a)) ≤∫ b

a

f ′(x)G(x)dx ≤M(f(b)− f(a)).

Since f(a) ≥ 0, we have

f(a)G(a) +

∫ b

a

f ′(x)G(x)dx ≤ f(a)M +M(f(b)− f(a)) = f(b)M.

Similarly,

f(a)G(a) +

∫ b

a

f ′(x)G(x)dx ≥ f(b)m.

Therefore,

mf(b) ≤∫ b

a

f(x)g(x)dx ≤Mf(b).

If f(b) = 0, take any ξ, we have the proof. If f(b) �= 0, then f(b) > 0, we have

m ≤ 1

f(b)

∫ b

a

f(x)g(x)dx ≤M.

Since G(x) is continuous, by the Intermediate Value Theorem, there exists a ξ ∈ [a, b], such that

1

f(b)

∫ b

a

f(x)g(x)dx = G(ξ) =

∫ b

ξ

g(x)dx,

or ∫ b

a

f(x)g(x)dx = f(b)

∫ b

ξ

g(x)dx.

Case (b) can be proved similarly by introducing

G(x) =

∫ x

a

g(t)dt.

For case (c), if f ′(x) ≥ 0, let F (x) = f(x)− f(a), then F (x) satisfies the conditions in case (a).

The conditions in the above example can be relaxed, and we have the following more general

Second Mean Value Theorem for Integrals:

Suppose that g(x) is integrable on [a, b], then we have the following results:

(a) If f(x) is monotonically increasing on [a, b] and f(x) ≥ 0, then there exists a ξ ∈ [a, b], s.t.,∫ b

a

f(x)g(x)dx = f(b)

∫ b

ξ

g(x)dx.


(b) If f(x) is monotonically decreasing on [a, b] and f(x) ≥ 0, then there exists a ξ ∈ [a, b], s.t.,

∫ b

a

f(x)g(x)dx = f(a)

∫ ξ

a

g(x)dx.

(c) In general, if f(x) is monotonic on [a, b], then there exists a ξ ∈ [a, b], s.t.,

∫ b

a

f(x)g(x)dx = f(a)

∫ ξ

a

g(x)dx + f(b)

∫ b

ξ

g(x)dx.

Theorem 6.4.8 (Integration by Substitution) Let the function f : [a, b] → R be continuous. Suppose that

the function g : [c, d] → R and its derivative g ′ are also continuous on [c, d], and, moreover, that g(c) = a,

g(d) = b, and a ≤ g(t) ≤ b ∀t ∈ [c, d]. Then,

∫ d

c

f(g(t))g′(t)dt =∫ b

a

f(x)dx.

Proof. Define H : [c, d] → R by

H(t) =

∫ t

c

f(g(s))g′(s)ds−∫ g(t)

g(c)

f(x)dx ∀t ∈ [c, d].

Since f(g(t)) is continuous on [c, d] by Continuity of Composition functions, then f(g(t))g ′(t) is con-

tinuous on [c, d]. Then by Proposition 6.4.4 H(t) is continuous on [c, d]. Moreover, from the Second

Fundamental Theorem and Corollary 6.4.6 we have

H ′(t) = f(g(t))g′(t)− f(g(t))g′(t) = 0 ∀t ∈ (c, d),

i.e., H(t) is a constant. Since H(t) is continuous and H(c) = 0, we get H(d) = 0.

Example 6.4.5∫ π

4

π6

xdx

1 + x4.

Solution.

∫ π4

π6

xdx

1 + x4=

∫ π4

π6

1

2

dx

1 + x4=

1

2

∫ (π4 )2

(π6 )2

dy

1 + y2

=1

2

[tan−1y

](π4 )2

(π6 )2

=1

2

[tan−1(

π

4)2 − tan−1(

π

6)2]

Example 6.4.6∫ 1

0

x2exdx.


Solution. By the method of Integration by Substition, it can be found that the antiderivative of f(x) = x 2ex

is

F (x) = ex[x2 − 2x+ 2]

By the theorem, ∫ 1

0

x2exdx = F (1)− F (0) = e− 2

You can also calculate the integral in the following way:∫ 1

0

x2exdx =

∫ 1

0

x2dex = [x2ex]10 −∫ 1

0

ex2xdx

= e− 2

∫ 1

0

xdex = e− 2[xex]10 + 2

∫ 1

0

exdx

= e− 2e+ 2[ex]10 = −e+ 2e− 2

= e− 2

6.5 Exercise

1. Consider the partition P = [0, 1/4, 1/2, 1] of the interval [0, 1]. Compute L(f, P ) and U(f, P ) for

f(x) = x, x ∈ [0, 1].

2. Let f(x) = mx+ b, x ∈ [0, 1], where m and b are positive.

(a) Show that the area under the graph of y = f(x) and above the x axis is b+m/2.

(b) For any partition P , show that L(f, P ) < b+m/2 < U(f, P ).

3. Suppose that the bounded function f : [a, b] → R has the property that for each rational number x

in the interval [a, b], f(x) = 0. Prove that I∗ ≤ 0 ≤ I∗.

4. Suppose that the bounded function f : [a, b] → R has the property that f(x) ≥ 0 ∀x ∈ [a, b]. Prove

that I∗ ≥ 0.

5. Suppose that the two bounded function f : [a, b] → R and g : [a, b] → R have the property that

g(x) ≤ f(x) ∀x ∈ [a, b].

(a) For P a partition of [a, b], show that L(g, P ) ≤ L(f, P ).

(b) Use part (a) to show that I∗(g) ≤ I∗(f).

6. Define

f(x) =

⎧⎨⎩ x, if x ∈ [0, 1] is rational,

0, if x ∈ [0, 1] is irrational.

Prove that I∗ = 0 and I∗ ≥ 1/2.


7. Let the function f : [a, b] → R be monotonically decreasing and let Pn be the regular partition of

[a, b] into n intervals of equal length (b − a)/n. Show that

U(f, Pn)− L(f, Pn) =[f(a)− f(b)](b− a)

n,

and then show that f is integrable on [a, b].

8. Define

f(x) =

⎧⎨⎩ x, if 2 ≤ x ≤ 3,

2, if 3 ≤ x ≤ 4.

Prove that f : [2, 4] → R is integrable.

9. Let

f(x) =

⎧⎨⎩ sgn

(sin π

x

), x �= 0,

0, x = 0.

Prove that f(x) is integrable on [0, 1].

10. Use the Integrability Criterion to show that f(x) = x2, x ∈ [0, 1] is integrable.

11. Define

f(x) =

⎧⎨⎩ x, if x ∈ [0, 1] is rational,

−x, if x ∈ [0, 1] is irrational.

Prove that f : [0, 1] → R is not integrable.

12. Let f(x) = x+1, x ∈ [0, 1], then f(x) is integrable since f(x) is monotonically increasing. Let Pn

be the regular partition of [0, 1] into n intervals of equal length 1/n. Compute U(f, P n) of f , and

then find∫ 10 f(x)dx by letting n→ ∞.

13. Suppose that the functions f, g, f 2, g2 and fg are integrable on the closed bounded interval [a, b].

Prove that [f − g]2 also is integrable on [a, b] and that∫ ba[f − g]2dx ≥ 0. Use this to prove that

∫ b

a

f(x)g(x)dx ≤ 1

2

[∫ b

a

f2(x)dx +

∫ b

a

g2(x)dx

].

14. Suppose that the continuous function f : [a, b] → R has∫ baf(x)dx = 0. Prove that there is some

point x0 ∈ [a, b] at which f(x0) = 0. Hint. Use the Extreme Value Theorem and the Intermediate Value Theorem.

15. Suppose the continuous function f : [a, b] → R has the property that∫ dc f(x)dx ≤ 0 ∀a ≤ c ≤ d ≤

b. Prove that f(x) ≤ 0 ∀x ∈ [a, b]. Is this true if we require only integrability of the function?

16. Suppose that the function f : [0, 1] → R is continuous and that f(x) ≥ 0 ∀x ∈ [0, 1]. Prove that∫ 10 f(x)dx > 0 if and only if there is a point x0 ∈ [0, 1] at which f(x0) > 0.


17. Suppose that the functions f : [a, b] → R and g : [a, b] → R are continuous. Prove that

∫ b

a

|f(x) + g(x)|dx ≤∫ b

a

|f(x)|dx +

∫ b

a

|g(x)|dx.

18. Suppose that the function f : [a, b] → R is Lipschitz, that is, that there is a constant c ≥ 0 such that

|f(u)−f(v)| ≤ c|u−v| ∀u, v ∈ [a, b]. For a partitionP of [a, b], prove that 0 ≤ U(f, P )−L(f, P ) ≤c|b− a|‖P‖.

19. Show that a Lipschitz function is integrable.

20. Suppose that the functions f, g, f 2, g2 and fg are integrable on the closed bounded interval [a, b].

prove that ∫ b

a

f(x)g(x)dx ≤(∫ b

a

f2(x)dx

)1/2(∫ b

a

g2(x)dx

)1/2

.

21. Suppose that the functions f : [a, b] → R is bounded and that it is continuous except at a finite

number of points. Prove that f : [a, b] → R is integrable.

22. Define

f(x) =

⎧⎨⎩ 1, if x ∈ [0, 1] is rational,

−1, if x ∈ [0, 1] is irrational.

Prove that f : [0, 1] → R is not integrable.

23. Suppose that the functions f(x) is integrable on [a, b]. Prove that |f(x)| is also integrable on [a, b].

Is the reverse true? (Consider the above problem.)

24. Suppose that the functions f : [a, b] → R is continuous and∫ b

a

f2(x)dx = 0. Prove that f(x) ≡ 0

on [a, b]. Hint. Prove by contradiction.

25. Suppose that the functions f : [−a, a] → R is continuous. Prove that∫ a

−a

f(x)dx = 0 if f(x) is an

odd function, and that∫ a

−a

f(x)dx = 2

∫ a

0

f(x)dx if f(x) is an even function.

26. Use the First Fundamental Theorem to evaluate each of the following integrals:

(a)∫ 2

1

[ 1x2

+ x+ cosx]dx.

(b)∫ 1

0

x√

4− x2dx.

(c)∫ 1

0

xdx

1 +√1 + x

.

(d)∫ π/2

0

cosn xdx.

(e)∫ 3

1

x√10− xdx.

(f)∫ π

0

cos2 xdx.

(g)∫ 1

0

(1− x2)ndx.


27. Find the following limits by using Riemann sums:

(a)

limn→∞

[1β + 2β + · · ·+ nβ

nβ+1

].

(b) limn→∞

[ n∑k=1

k

n2 + k2

].

(c) limn→∞

[ 1

n+ 1+

1

n+ 2+ · · ·+ 1

2n

].

(d) limn→∞

[ 1√n · n +

1√n · (n+ 1)

+ · · ·+ 1√n · (n+ n)

].

28. Find the area of the region bounded by:

(a) y = x2 and x+ y = 2.

(b)x2

a2+y2

b2, y = 0.

(c) y = x and y = x+ sin2 x, 0 ≤ x ≤ π.

29. A car runs on the road for one hour with the instant velocity

v(t) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

500t, 0 ≤ t ≤ 0.1,

50, 0.1 ≤ t ≤ 0.9,

500(1− t), 0.9 ≤ t ≤ 1,

where the unit is mile/hour. Find the distance that the car covers.

30. Calculate the following derivatives:

(a)d

dx

(∫ x

0

x2t2dt)

.

(b)d

dx

(∫ ex

1

ln tdt)

.

(c)d

dx

(∫ x

−x

et2

dt)

.

(d)d

dx

(∫ x

1

cos(x+ t)dt)

.

31. Suppose that the functions f : R → R is differentiable. Define the function H : R → R by

H(x) =

∫ x

−x

[f(t) + f(−t)]dt ∀x. Find H ′′(x).

32. Suppose that the function f : R → R is continuous. Define G(x) =∫ x

0

(x− t)f(t)dt ∀x. Prove

that G′′(x) = f(x) ∀x.

33. For numbers a1, · · · , an, define p(x) = a1x+a2x2+ · · ·+anxn for all x. Suppose that

a12+a23+

· · ·+ ann+ 1

= 0. Prove that there is some point x ∈ (0, 1) such that p(x) = 0.

34. Suppose that the function f : R → R has a continuous second derivative. Prove that

f(x) = f(0) + f ′(0)x+

∫ x

0

(x− t)f ′′(t)dt ∀x.

35. Suppose that the functions f : [a, b] → R and g : [a, b] → R are bounded and integrable. Prove

that fg is integrable.

36. Suppose that f : [a, b] → R is continuous, and∫ b

a

f(x)g(x)dx = 0

for any continuous function g(x) satisfying g(a) = g(b) = 0. Prove that f(x) ≡ 0.

Hint. If not, ∃x0 ∈ (a, b), s.t. (e.g.), f(x0) > 0, ∀x ∈ (x0 − δ, x0 + δ), then take a proper g(x).

Chapter 7: Series of Numbers 139

37. Suppose that f : [a, b] → R is continuous and satisfies the conditions

∫ b

a

xkf(x)dx = 0, k = 0, 1, · · · , n.

Prove that the equation f(x) = 0 has at least n+ 1 solutions in (a, b). Hint. By induction.

38. Suppose that f : [a, b] → R is integrable. Show that for any [c, d] ⊂ [a, b], there exists x0 ∈ [c, d],

such that f(x) is continuous at x0.

Hint. Let ω(c, d) = supx∈[c,d]

f(x) − infx∈[c,d]

f(x), then for given ε > 0, ∃[c, d] ⊆ [a, b], s.t., ω(c, d) < ε.

Otherwise, f(x) is not integral on [a, b]. Let a1 = a, b1 = b, then ∃[a2, b2] ⊆ [a1, b1], s.t., ω(a2, b2) <1/2, · · · , ω(an, bn) < 1/n, By Nested Interval Theorem, ∃c ∈ [an, bn]. For given ε > 0, take n s.t., 1/n < ε,and take δ s.t., (c− δ, c+ δ) ⊆ [an, bn], then |f(x) − f(c)| ≤ ω(an, bn) < 1/n < ε, ∀x ∈ (c− δ, c+ δ), i.e.,f(x) is continuous at c.

39. (a). Suppose that f(x) and f ′(x) are continuous on [a, b]. Prove that

limn→∞

∫ b

a

f(x) sinnxdx = 0.

(b). Suppose that f(x) is integrable on [0, 2π]. Prove that

limn→∞

∫ 2π

0

f(x) sinnxdx = 0.

Hint. (a). Integration by parts. (b).∫ b

af(x) sinnxdx =

n∑k=1

∫ (2k−1)πn

(2k−2)πn

f(x) sinnxdx+n∑

k=1

∫ 2kπn

(2k−1)πn

f(x) sinnxdx =n∑

k=1

2

nλk −

n∑k=1

2

nμk ,

where mk ≤ λk , μk ≤ Mk , and mk = minx∈[

(2k−2)πn

, 2kπn

]

f(x), Mk = maxx∈[

(2k−2)πn

, 2kπn

]

f(x).

Since f(x) is integral, limn→∞

∫ 2π

0f(x) sinnxdx = lim

n→∞1

π

[ n∑k=1

λk2π

n−

n∑k=1

μk2π

n

]= 0.


Chapter 7

Series of Numbers

7.1 Definition and Basic Properties

Definition. Let {ak} be a sequence of real numbers. Then the sum∞∑k=1

ak is called a Series (or Number

Series), ak is called the kth term of the series. The sum

Sn =n∑

k=1

ak

for any natural number n is called a Partial Sum. If the sequence of partial sums {S n} converges, i.e.,

limn→∞Sn = S,

then we say that the series converges and call S the sum of the series, denoted by

∞∑k=1

ak = S.

If {Sn} does not converge, then we say that the series∞∑k=1

ak diverges.

Proposition 7.1.1 (necessary condition for convergence) Suppose that the series∞∑k=1

ak converges. Then

limk→∞

ak = 0.

Proof. The convergence of∞∑k=1

ak means that limn→0

Sn = S. Hence

limk→∞

ak = limn→∞(Sk − Sk−1)

= limn→∞Sk − lim

n→∞Sk−1 = S − S = 0.

141


We have learnt the Cauchy Convergence Criterion for sequence. Apply to Series, we obtain the follow-

ing proposition.

Proposition 7.1.2 (Cauchy Convergence Criterion) A series∞∑k=1

ak converges if and only if for every

ε > 0, there exists a natural number N such that

|an+1 + an+2 + · · ·+ am| < ε, if m > n ≥ N.

Proof. Using Cauchy Convergence Criterion to the sequence of partial sums {S n}, {Sn} converges if and

only if for every ε > 0, there exists a natural numberN such that

|Sm − Sn| < ε, if m > n ≥ N.

But

|Sm − Sn| = |an+1 + · · ·+ am|,

i.e.,

|an+1 + · · ·+ am| < ε, if m > n ≥ N.

Proposition 7.1.3 (Convergence of Geometric Series) If |r| < 1, then

∞∑k=0

rk =1

1− r.

Proof. The partial sum of the series is

Sn =

n∑k=0

rk =1− rn+1

1− r.

Hence ∞∑k=0

rk = limn→∞Sn =

1

1− r.

If |r| ≥ 1, the general terms ak = rk don’t converge to 0. Hence the series∞∑k=0

rk does not converge. The

Geometric Series is the simplest series but will play an important role in the following Comparison Test.

7.2 Series with Nonnegative Terms

Definition 7.2.1 If ak ≥ 0 for all k, then∞∑k=1

ak is called a series with nonnegative terms.


Since ak ≥ 0 for all k, the partial sums satisfy

Sn = Sn−1 + an ≥ Sn−1 for all n.

The sequence {Sn} is monotonically increasing. Hence by the limit theory, we have the following proposi-

tion.

Proposition 7.2.2 A series with nonnegative terms converge if and only if the sequence of partial sums {S n}has an upper bound.

Proof. It’s known that {Sn} is monotonically increasing. By the Monotone Convergence Theorem, {S n}converges if {Sn} has an upper bound. On the other hand, we know that a convergent sequence must be

bounded.

Theorem 7.2.3 (Comparison Test) Suppose that {ak} and {bk} are sequences of numbers such that 0 ≤ak ≤ bk ∀k.

(i) If∞∑k=1

bk converges, then∞∑n=1

ak converges.

(ii) If∞∑

n=1

ak diverges, then∞∑

n=1

bk diverges.

Proof. Let {An} and {Bn} be sequences of partial sums of∞∑k=1

ak and∞∑k=1

bk respectively. From the

assumption 0 ≤ ak ≤ bk ∀k, we know

An ≤ Bn for all n.

Hence, by the above Proposition, it follows

(i) If∞∑k=1

bk converges, then {Bn} is bounded above. From An ≤ Bn, {An} is bounded above. There-

fore∞∑k=1

ak converges.

(ii) If∞∑k=1

ak diverges, then {An} has no upper bound. From An ≤ Bn, {Bn} has no upper bound also.

Therefore∞∑k=1

bk diverges.

Corollary 7.2.4 Let N be a natural number and C is a positive real number. Suppose that

0 ≤ an ≤ Cbn for n ≥ N.

(i) If∞∑k=1

bk converges, then∞∑k=1

ak converges.


(ii) If∞∑k=1

ak diverges, then∞∑k=1

bk diverges.

Proof. It’s obvious that∞∑k=1

bk converges if and only if∞∑

k=N

bk converges, where N is any natural number.

Also,∞∑k=1

bk converges if and only if∞∑k=1

Cbk converges, where C is a nonzero constant. Hence∞∑k=1

bk con-

verges,∞∑k=1

Cbk converges and∞∑

k=N

Cbk converges.∞∑

k=N

ak and∞∑

k=N

Cbk are two series with nonnegative

terms, by Comparison Test,∞∑

k=N

ak converges due to convergence of∞∑

k=N

Cbk. Hence∞∑k=1

ak converges.

The proof of part (ii) is left as exercise.

Corollary 7.2.4’ Suppose that

limn→∞

anbn

= l.

Then,

(i) If 0 < j < +∞, then∞∑k=1

ak and∞∑k=1

bk converge the the same time, or diverge the same time.

(ii) If l = 0 and∞∑k=1

bk converges, then∞∑k=1

ak converges.

(iii) If l = +∞ and∞∑k=1

bk diverges, then∞∑k=1

ak diverges.

Theorem 7.2.5 (Ratio Test) Suppose that∞∑k=1

uk is a series with uk > 0 and satisfies

limk→∞

uk+1

uk= l.

(i) If l < 1, then the series converges.

(ii) If l > 1, then the series diverges.

(iii) If l = 1, it is indefinite.

Proof.

(i) Take a constant q such that l < q < 1. From the assumption limk→∞

uk+1

uk= l, there is a natural

numberN such that

uk+1

uk< q if k ≥ N.


It follows that

uN+1 < quN

uN+2 < quN+1 < q2uN

· · ·uN+k < · · · < qkuN

Since∞∑k=1

qkuN converges, by comparison test∞∑k=1

uN+k converges and then∞∑k=1

uk =

N∑k=1

uk +

∞∑k=1

uN+k converges.

(ii) Since limk→∞

uk+1

uk= l > 1, there is a natural numberN such that

un+1

un> 1, if n ≥ N,

i.e. uk+1 > uk. It means that {uk}k≥N is an increasing sequence and can not converge to 0. Hence

the series∞∑k=1

uk diverges.

(iii) Consider the series∞∑k=1

1

kp, p is a constant. It is known that when p = 2 the series converges and

p = 1 the series diverges. But for any real number p

limn→∞

1(n+1)p

1np

= 1.

Therefore, when l = 1, the ratio test is failed to give any conclusion.

Theorem 7.2.6 (Cauchy Test) Suppose that∞∑k=1

uk is a series with uk > 0 and satisfies

limk→∞

k√uk = l.

(i) If l < 1, then the series converges.

(ii) If l > 1, then the series diverges.

(iii) If l = 1, it is indefinite.

Proof.

(i) Take a constant q such that l < q < 1. From the assumption limk→∞

k√uk = l, there is a natural

numberN such that k√uk < q with k ≥ N , i.e.,

uk < qk, if k ≥ N.

By comparison test∞∑

k=N

uk converges due to the convergence of∞∑

k=N

qk, and then∞∑k=1

uk converges.


(ii) From the assumption limk→∞

k√uk = l > 1, there is a natural numberN such that

uk > 1, if k ≥ N.

{uk} can’t converge to 0, hence∞∑k=1

uk does not converge.

(iii) Also for example∞∑k=1

uk with uk =1

kp, limk→∞

k√uk = 1. But the series converges when p = 2 and

diverges when p = 1.

Definition 7.2.7 Suppose that f(x) is defined on [a,+∞) and for any A > a, f(x) is integrable in [a,A].

If the limit

limA→∞

∫ A

a

f(x)dx

exists, then we say that the integral∫ ∞

a

f(x)dx converges and denote by

∫ ∞

a

f(x)dx = limA→∞

∫ A

a

f(x)dx.

Proposition 7.2.8 Suppose that f(x) is nonnegative on interval [a,+∞) and for any A > a, f(x) is

integrable on [a,A]. Then∫ ∞

a

f(x)dx converges if and only if the function

F (A) =

∫ A

a

f(x)dx

has an upper bound.

Proof. Since f(x) is nonnegative, for A2 > A1 > a

F (A2)− F (A1) =

∫ A2

A1

f(x)dx ≥ 0.

Hence F (A) is monotonically increasing. From the limit theory, limA→∞

F (A) exist if and only if F(A) has

an upper bound.

Theorem 7.2.9 (Integral Test) Let∞∑k=1

uk be a series with nonnegative terms. Suppose that the function

f(x) is monotonically decreasing and satisfies

f(k) = uk.

Then the series∞∑k=1

uk converges if and only if∫ ∞

1

f(x)dx converges.

Proof. From the assumptions, we have

uk = f(k) ≥ f(x) ≥ f(k + 1) = uk+1 if k ≤ x ≤ k + 1.


Then we have

uk ≥∫ k+1

k

f(x)dx ≥ uk+1

andn∑

k=1

uk ≥∫ n+1

1

f(x)dx ≥n∑

k=1

uk+1,

or

Sn ≥∫ n+1

1

f(x)dx ≥ Sn+1 − u1.

If∫ ∞

1

f(x)dx converges, {Sn} has an upper bound∫ ∞

1

f(x)dx + u1. Therefore {Sn} converges. If∫ ∞

1

f(x)dx diverges, that means

{∫ n+1

1

f(x)dx

}∞

n=1

has no upper bound. By the inequality so {Sn}

has no upper bound, i.e.,∞∑k=1

uk diverges.

Using the Integral Test we get

Theorem 7.2.10 (p-Test) For a positive number p, the series∞∑k=1

1

kp

converges if and only if p > 1.

Proof. Consider the integral

∫ ∞

1

1

xpdx = lim

A→∞

∫ A

1

1

xpdx =

⎧⎪⎨⎪⎩

limA→∞

1

1− p(A1−p − 1), p �= 1

limA→∞

lnA, p = 1

The integral converges only for p > 1.

Example 7.2.1 Consider the convergence of the following series:

(a)∞∑

n=2

1

n lnn.

(b)∞∑

n=2

1

(lnn)lnn.

(c)∞∑n=1

(1000)n

n!.

(d)∞∑n=1

n1000

2n.

7.3 Convergence of General Series

Now we consider the general series∞∑k=1

uk, i.e., uk may be positive or negative. But first we discuss

the Alternating Series.

Theorem 7.3.1 (Alternating Series Test) Suppose that {uk} is a monotonically decreasing sequence of

nonnegative numbers and limk→∞

uk = 0. Then the series

∞∑n=1

(−1)n+1un


converges.

Proof. We first show that the subsequences {S2n} and {S2n+1} converge. Since {uk} is monotonically

decreasing,

S2n+2 − S2n = u2n+1 − u2n+2 ≥ 0

and

S2n = u1 −n−1∑k=1

(u2k − u2k+1)− u2n ≤ u1.

We conclude that {S2n} is monotonically increasing and bounded by u 1. Hence {S2n} converges. But

S2n+1 = S2n + u2n+1

and limn→∞un = 0, then lim

n→∞S2n+1 = limn→∞S2n. Denote this limit by S. Now we show that the sequence

{Sn} converges to S. For given ε > 0, we choose natural numbersN 1 and N2 such that

|S2n − S| < ε if n ≥ N1

and

|S2n+1 − S| < ε if n ≥ N2.

Define N = max{2N1, 2N2 + 1}. Then

|Sn − S| < ε if n ≥ N.

Example 7.3.1 Discuss the convergence of the following series:

1 +1

2+

1

3− 1

4− 1

5− 1

6+

1

7+

1

8+

1

9− · · · .

Definition 7.3.2 A series∞∑k=1

uk is said to converge absolutely provided that the series∞∑k=1

|uk| converges.

Theorem 7.3.3 (Absolute Convergence Test) The series∞∑k=1

uk converges if the series converges absolute-

ly, i.e.∞∑k=1

|uk| converges.

Proof. Since∞∑k=1

|uk| converges, by using Cauchy Convergence Criterion to series, for given ε > 0, ∃Nsuch that

|un+1|+ |un+2|+ · · ·+ |un| < ε, if m > n ≥ N,

and then

|un+1 + un+2 + · · ·+ un| ≤ |un+1|+ · · ·+ |um| < ε.


Hence again by Cauchy Convergence Criterion,∞∑k=1

uk converge.

Note.∞∑k=1

uk converges,∞∑k=1

|uk| may not converge (Consider∞∑k=1

(−1)k+1

k).

According to this theorem, all the convergence tests mentioned in Section 2 for series with nonnegative

terms can be used as sufficient conditions for general series.

Example 7.3.2 Consider the series∞∑n=1

sinn

n2. Since

∣∣∣∣ sinnn2

∣∣∣∣ ≤ 1

n2

and∞∑n=1

1

n2converges. Then

∞∑n=1

∣∣∣∣ sinnn2

∣∣∣∣ converges, i.e.,∞∑n=1

sinn

n2converges absolutely. By Theorem 7.3.3,

∞∑n=1

sinn

n2converges.

Definition 7.3.4 (Conditional Convergence) A series that converges but does not converge absolutely is

said to converge conditionally.

Example 7.3.3∞∑n=1

(−1)n+1

nconverges conditionally.

There are some interesting properties related to absolute convergence and conditional convergence.

For examples:

• If a series converges absolutely, then rearranging the terms in any order the new series still converges

absolutely and the sum does not change.

• If a series converges conditionally, then we can rearrange the terms in some order to make the partial

sums sequence converge to any specified number including +∞ and −∞.

Next, we introduce two tests: Abel and Dirichlet Tests, which are very useful for some series. Consider

the partial sum:

S =

m∑k=1

akbk.

Let

Bj =

j∑k=1

bk, j = 1, 2, · · · ,m,

then

b1 = B1, bk = Bk −Bk−1, k = 2, 3, · · · ,m.


Let B0 = 0, then

S =

m∑k=1

akbk =

m∑k=1

ak(Bk −Bk−1)

=

m∑k=1

akBk −m∑

k=2

akBk−1

= amBm +

m−1∑k=1

akBk −m−1∑k=2

ak+1Bk

= amBm +m−1∑k=1

(ak − ak+1)Bk.

Lemma 7.3.5 (Abel) If {a1, a2, · · · , am} and {b1, b2, · · · , bm} satisfy the conditions:

(i) {ak} is monotone;

(ii) {Bk} is bounded, i.e., ∃M > 0, s.t.,

|Bk| ≤M, k = 1, 2, · · · ,m.

Then we have

|S| =∣∣∣ m∑k=1

akbk

∣∣∣ ≤M(|a1|+ 2|am|).

Proof. From the expression of S, we get

|S| ≤m−1∑k=1

|ak − ak+1| · |Bk|+ |amBm|

≤ M[m−1∑

k=1

|ak − ak+1|+ |am|].

Since {ak} is monotone, we have

|S| ≤ M[|a1 − am|+ |am|

]≤ M(|a1|+ 2|am|).

Theorem 7.3.6 (Abel Test) Suppose the the series∞∑

n=1

anbn satisfies

(i)∞∑n=1

bn converges;

(ii) {an} is monotone and bounded.

Then∞∑

n=1

anbn converges.


Proof. From (ii), there is a M such that |an| ≤ M, ∀n. Since∞∑

n=1

bn converges, then for given ε > 0, ∃N ,

s.t.,∣∣∣ n+p∑k=n+1

bk

∣∣∣ < ε

3M, ∀p > 0, n > N. From Abel’s lemma we get

∣∣∣ n+p∑k=n+1

akbk

∣∣∣ ≤ ε

3M(|an+1|+ 2|an+p|) ≤ ε.

By Cauchy Convergence Criterion,∞∑

n=1

anbn converges.

Theorem 7.3.7 (Dirichlet Test) Suppose the the series∞∑n=1

anbn satisfies

(i) Bn =

n∑k=1

bk is bounded;

(ii) {an} converges to zero monotonically.

Then∞∑

n=1

anbn converges.

Proof. Suppose that |Bn| ≤M, ∀n, then

|bn+1 + · · ·+ bn+p| = |Bn+p −Bn| ≤ 2M, ∀n, p.

Since {an} → 0, then for given ε > 0, ∃N , s.t., |an| < ε

6M, ∀n > N . From Abel’s lemma we have

∣∣∣ n+p∑k=n+1

akbk

∣∣∣ ≤ 2M(|an+1|+ 2|an+p|) ≤ ε, ∀p > 0, n > N.

By Cauchy Convergence Criterion,∞∑

n=1

anbn converges.

7.4 Exercise

1. Examine the following series for convergence:

(a)∞∑

n=1

sin2 nθ

n2 + 1

(b)∞∑

n=1

2 · 5 · 8 · · · (3n− 1)

1 · 5 · 9 · · · (4n− 3)

(c)∞∑

n=1

2n sinx

3n, x > 0

(d)∞∑k=1

( k + 1

k2 + 1

)3

(e)∞∑n=1

1000n

n!

(f)∞∑k=2

1

(ln k)ln k

2. Examine the following series for convergence and decide whether the series converges absolutely:


(a)∞∑k=1

(−1)kck

k, c > 1

(b)∞∑k=2

(−1)k1

ln k

(c)∞∑

n=1

cosnx

np, p > 0, 0 < x < π.

(d)∞∑k=1

(−1)kk5

k!

(e)∞∑k=2

(ln k)(sin(kπ/3))

k2

3. For any positive number α, prove that the series

∞∑k=1

kα

ek

converges.

4. Fix a positive number α and consider the series

∞∑k=1

1

(k + 1)[ln(k + 1)]α.

For what values of α does this series converge?

5. (The Cauchy Root Test) For the series∞∑k=1

ak, suppose that there is a number r with 0 ≤ r < 1 and

a naturalN such that

|ak|1/k < r ∀k ≥ N.

Prove that∞∑k=1

ak converges absolutely.

6. Use the Cauchy Convergence Criterion for series to provide another proof of the Alternating Series

Test.

7. Let

u+n =

⎧⎨⎩ un, un ≥ 0,

0, un < 0,u−n =

⎧⎨⎩ 0, un ≥ 0,

−un, un < 0.

Show that (i).∞∑

n=1

un converges absolutely if and only if both∞∑

n=1

u+n and∞∑n=1

u−n converge. (ii). If

∞∑n=1

un converges conditionally, then both∞∑n=1

u+n and∞∑n=1

u−n diverge.

8. Prove that if a series converges absolutely, then we can rearrange the terms of the series in any order,

the new series still converges absolutely and the sum is the same.

Chapter 8

Series of Functions

8.1 Uniform Convergence

8.1.1 Pointwise Convergence

Definition 8.1.1 A sequence of functions {fn(x)} defined on the interval I is said to converge pointwise (or

simply converge) to f(x) on I , if for each x0 ∈ I ,

limn→∞ fn(x0) = f(x0).

Definition 8.1.1′ A series of functions∞∑n=1

un(x) defined on the interval I is said to converge pointwise to

S(x) on I , if for each x0 ∈ I , the sequence of partial sums {Sn(x)}, where Sn(x) =n∑

k=1

uk(x), satisfies

limn→∞Sn(x0) = S(x0),

i.e. for every point of I , Sn(x) converges to S(x).

In many problems of mathematics, we may need to know the properties of the limit function. In other

words, we may need the answers to the following questions:

(1) If {fn(x)} are all continuous on the interval I , and converges pointwise to f(x). Is f(x) continuous?

(2) If {fn(x)} are all differentiable on the interval I and converges pointwise to f(x). Is f(x) differen-

tiable and limn→∞ f ′

n(x) = f ′(x)?

(3) If {fn(x)} are all integrable on the interval [a, b] and converges pointwise to f(x). Is f(x) integrable

153


and

limn→∞

∫ b

a

fn(x)dx =

∫ b

a

f(x)dx.

Example 8.1.1 Let fn(x) = xn, 0 ≤ x ≤ 1. Then

limn→∞ xn = f(x) =

⎧⎨⎩ 0, x �= 1,

1, x = 1.

For any n, fn(x) are all continuous on [0, 1] and differentiable but the limit function f(x) is not continuous

and not differentiable on [0, 1].

Example 8.1.2 Let fn(x) = nxe−nx2

, 0 ≤ x ≤ 1. Then

limn→∞ fn(x) = 0 ∀x ∈ [0, 1].

The integral of fn is ∫ 1

0

fn(x)dx =

∫ 1

0

e−nx2

dnx2

2

=1

2

(−enx2

)∣∣∣10=

12 − e−n

2.

Then we have

limn→∞

∫ 1

0

fn(x)dx =1

2.

However, since∫ 1

0

[ limn→∞ fn(x)] = 0, then

limn→∞

∫ 1

0

fn(x)dx �=∫ 1

0

[ limn→∞ fn(x)]dx.

The above examples show that the assumption of Pointwise Convergence can not guarantee affirmative

answers to the above three questions. To find the answers to these questions, we need the concept of uniform

convergence.

8.1.2 Uniform Convergence

Definition 8.1.2 A sequence of functions {fn(x)} is said to converge uniformly to f(x) on the interval I , if

for any given ε > 0, there exists a natural number N (which may depend on ε but not x.) such that for all

x ∈ I

|fn(x) − f(x)| < ε, if n ≥ N.

Chapter 8: Series of Functions 155

The geometrical interpretation is that the graph of functions fn(x), n ≥ N , lie between the graph of

function f(x)− ε and function f(x) + ε.

Example 8.1.3 Let fn(x) = xn. Then {fn(x)} converges uniformly to f(x) = 0 on the interval [0, 1− δ],

0 < δ < 1, but does not converge uniformly to f(x) = 0 on the interval [0, 1].

Proof. For any ε > 0, choose N =

[ln ε

ln(1− δ)

]+ 1, then for all x ∈ [0, 1− δ],

|fn(x)− 0| = |xn| ≤ (1− δ)n < ε , if n ≥ N

Hence {fn(x)} converges uniformly to f(x) = 0 on [0, 1− δ]. On the interval [0, 1],

limn→∞ fn(x) = f(x) =

⎧⎨⎩ 0, 0 ≤ x < 1,

1, x = 1,

We now prove that {fn(x)} does not converge uniformly on [0, 1]. We argue by contradiction. If {f n(x)}converges uniformly on [0, 1], then when for ε = 1/2, there is a natural numberN such that

|fn(x) − f(x)| < 1

2if n ≥ N.

Thus for this N , we have

|xN − 0| < 1

2for 0 ≤ x < 1

But it is obvious that limx→1

xN = 1, which means that fN (x) can be arbitrarily close to 1 as x sufficiently

close to 1. For example, choose x0 = (2/3)1/N , then 0 ≤ x0 < 1, and

xN0 =2

3>

1

2.

Thus we get a contradiction. It shows that f(x) can not converge uniformly on [0, 1].

Definition 8.1.3 A function series∞∑n=1

un(x) defined on the interval I is said to converge uniformly to S(x)

on the interval I , if for any given ε > 0, there exists a natural numberN such that for all x ∈ I , the partial

sum Sn(x) =n∑

k=1

uk(x) satisfies

|Sn(x)− S(x)| < ε , if n ≥ N.

8.1.3 Test for Uniform Convergence

Theorem 8.1.4 (Cauchy Convergence Criterion) A sequence of functions {fn(x)} converges uniformly

on the interval I , if and only if for any given ε > 0, there exists a natural numberN such that for all x ∈ I

|fm(x)− fn(x)| < ε, if m > n ≥ N.


Proof. If {fn(x)} converges to f(x) uniformly, then for given ε > 0, ∃N , s.t.,

|fn(x)− f(x)| < ε

2, ∀n ≥ N, x ∈ I.

Then,

|fm(x)− fn(x)| < |fm(x) − f(x)|+ |fn(x)− f(x)| < ε, ∀m,n ≥ N, x ∈ I.

If {fn(x)} is a Cauchy sequence, then {fn(x)} converges pointwise to f(x), then for given ε > 0, ∃N , s.t.,

|fm(x)− fn(x)| < ε, ∀m,n ≥ N, x ∈ I.

Fix n, and let m→ +∞, we get

|fn(x)− f(x)| < ε, ∀n ≥ N, x ∈ I,

i.e., {fn(x)} converges uniformly on the interval I .

Theorem 8.1.4′ (Cauchy Convergence Criterion) A function series∞∑n=1

un(x) converges uniformly on the

interval I , if and only if for any given ε > 0 there exists a natural numberN , such that for all x ∈ I

|un+1(x) + · · ·+ um(x)| < ε, if m > n ≥ N.

Theorem 8.1.5 (Weierstrass Uniform Convergence Criterion) If the general term un(x) of∞∑

n=1

un(x)

satisfy

|un(x)| ≤ an, ∀x ∈ I

and the series∞∑

n=1

an converges. Then the function series∞∑n=1

un(x) converges uniformly on the interval I .

Proof. Since∞∑n=1

an converges, by Cauchy Convergence Criterion, for given ε > 0, ∃N such that

|an+1 + · · ·+ am| < ε, if m > n ≥ N.

But ak are all nonnegative,

|an+1 + · · ·+ am| = an+1 + · · ·+ am < ε.

From the assumption |un(x)| ≤ an, ∀x ∈ I , if follows that for all x ∈ I ,

|un+1(x) + · · ·+ um(x)| ≤ |un+1(x)|+ · · ·+ |um(x)|≤ an+1 + · · ·+ am < ε, if m > n ≥ N.


By Theorem 8.1.4,∞∑

n=1

un(x) converges uniformly on I .

According to this theorem, all the tests for number series with nonnegative terms can be used for the test of

uniform convergence of function series.

Example 8.1.4 Consider the function series

∞∑n=1

sin(nπx)

np, −∞ < x < +∞, p > 1.

Since ∣∣∣∣sin(nπx)np

∣∣∣∣ ≤ 1

np, −∞ < x < +∞

and∞∑n=1

1

npconverges when p > 1, the function series converges uniformly on (−∞,+∞).

Theorem 8.1.6 (Dini Test) Suppose the {fn(x)} converges pointwise to f(x) on [a, b], fn(x), ∀n, and f(x)

are continuous on [a, b], and for any x ∈ [a, b] {fn(x)} is monotone. Then {fn(x)} converges uniformly to

f(x) on [a, b].

Proof. By contradiction. If {fn(x)} does not converge uniformly to f(x) on [a, b], then ∃ε 0 > 0, for any

nk, ∃xk, s.t.,

|fnk(xk)− f(xk)| ≥ ε0, k = 1, 2, · · · ,

Since {xk} is bounded, then {xk} has a converged subsequence {xki}, xki → x0 ∈ [a, b], as i → +∞.

Since fn(x0) → f(x0), ∃N , s.t.,

|fN (x0)− f(x0)| < ε0.

Since fN (x)− f(x) is continuous at x0, we have

limi→+∞

|fN (xki)− f(xki)| = |fN (x0)− f(x0)| < ε0,

and ∃I , s.t.,

|fN(xki )− f(xki)| < ε0, ∀i > I.

For each x, {fn(x)} is monotone, then

|fnk(xki )− f(xki)| ≤ |fN(xki )− f(xki)| < ε0, ∀i > I, nk > N,

which is a contradiction.

Example 8.1.5 Suppose that f(x) is continuous on [0, 1] and f(1) = 0. Show that g n(x) = f(x)xn, n =

1, 2, · · · , converges uniformly on [0, 1].

Proof. gn(x) and g(x) are continuous on [0, 1]. For each x ∈ [0, 1], {xn} is monotone, then {gn(x)} is

monotone. By Dini’s test, {gn(x)} converges uniformly on [0, 1].


Theorem 8.1.6’ (Series form of Dini Test) Suppose the∞∑n=1

un converges pointwise to S(x) on [a, b],

un(x), ∀n, and S(x) are continuous on [a, b], and for any x ∈ [a, b] all un(x) have the same sign. Then∞∑n=1

un converges uniformly to S(x) on [a, b].

Theorem 8.1.7 (Abel Test) Suppose that

(i) the function series∞∑

n=1

an(x) converges uniformly on the interval I;

(ii) the sequence {bn(x)} is monotone for given x ∈ I;

(iii) the sequence {bn(x)} is uniformly bounded on I , i.e., ∃M > 0, s.t.

|bn(x)| ≤M ∀x ∈ I, ∀n.

Then the function series∞∑n=1

an(x)bn(x) converges uniformly on I .

Proof. From (i) we have, for given ε > 0, ∃N , s.t.,

∣∣∣ n+p∑k=n+1

ak(x)∣∣∣ < ε

3M, ∀x ∈ I, ∀p > 0, n > N.

Since for each x ∈ I , {bn(x)} is monotone, from Abel’s lemma and (iii) we get

∣∣∣ n+p∑k=n+1

ak(x)bk(x)∣∣∣ ≤ ε

3M(|bn+1(x)| + 2|bn+p(x)|)

≤ ε

3M(M + 2M) = ε, ∀x ∈ I, ∀p > 0, n > N,

i.e.,∞∑n=1

an(x)bn(x) converges uniformly on I by Cauchy Convergence Criterion.

Example 8.1.6 Prove that the series∞∑

n=1

(−1)n√n

· 2 + xn

1 + xn

converges uniformly on [0,∞).

Proof. Let an(x) =(−1)n√

nand bn(x) =

2 + xn

1 + xn. Since

∞∑n=1

an(x) converges, then it converges uniformly.

For x ∈ [0, 1], {bn(x)} is increasing. For x ∈ (1,+∞), {bn(x)} is decreasing. Also,

2 + xn

1 + xn≤ 2 + 2xn

1 + xn= 2, ∀x ∈ [0,+∞), ∀n.

By Abel’s test,∞∑n=1

an(x)bn(x) converges uniformly on [0,∞).

Theorem 8.1.8 (Dirichlet Test) Suppose that


(i) An(x) =

n∑k=1

ak(x), n = 1, 2, · · · , is uniformly bounded on I , i.e., ∃M > 0, s.t.

|An(x)| ≤M ∀x ∈ I, ∀n;

(ii) for each x ∈ I , the sequence {bn(x)} is monotone;

(iii) the sequence {bn(x)} converges to 0 uniformly on I .

Then the function series∞∑n=1

an(x)bn(x) converges uniformly on I .

Proof. From (i) we have,

∣∣∣ n+p∑k=n+1

ak(x)∣∣∣ = |An+p(x)−An(x)| ≤ 2M, ∀x ∈ I, ∀p > 0, ∀n.

From (iii) we have, for given ε > 0, ∃N , s.t.,

|bn(x)| < ε

6M, ∀x ∈ I, ∀n > N.

From (ii) and Abel’s lemma we have,

∣∣∣ n+p∑k=n+1

ak(x)bk(x)∣∣∣ ≤ 2M(|bn+1(x)| + 2|bn+p(x)|)

≤ 2M( ε

6M+

2ε

6M

)= ε, ∀x ∈ I, ∀p > 0, ∀n > N,

i.e.,∞∑n=1

an(x)bn(x) converges uniformly on I by Cauchy Convergence Criterion.

Example 8.1.7 Prove that the series∞∑n=1

(−1)n(1− x)xn

converges uniformly on [0, 1].

Proof. Let an(x) = (−1)n and bn(x) = (1 − x)xn. An(x) =

n∑n=1

an(x) is uniformly bounded. Since

b′n(x) = nxn−1 − (n+ 1)xn, bn(0) = bn(1) = 0, then bn(x) reaches its maximum at xn =n

n+ 1. Then,

bn(x) <1

n+ 1, i.e., {bn(x)} converges to 0 uniformly on [0, 1]. Also, for each x ∈ [0, 1], {bn(x)} is

decreasing. By Dirichlet’s test,∞∑n=1

an(x)bn(x) converges uniformly on [0, 1].

8.2 Uniform Limit of Functions

With the concept of uniform convergence, we can now give answers to the three questions proposed in

Section 1.1.


8.2.1 Uniform Limit of Continuous Functions

Theorem 8.2.1 Suppose that {fn(x)} is a sequence of continuous functions defined on the interval I and

uniformly converges to f(x) in I , then the limit function f(x) is continuous in I .

Proof. Let x0 be any point I . We now prove ∀ε > 0, ∃δ > 0 such that

|f(x)− f(x0)| < ε if |x− x0| < δ.

Since fn(x) converges uniformly to f(x), there exists a natural numberN 0 such that

|fN0(x) − f(x)| < ε

3, ∀x ∈ I.

Because fN0(x) is continuous in interval I , especially continuous at the point x0, we can find a δ > 0 such

that

|fN0(x) − fN0(x0)| <ε

3, if |x− x0| < δ.

Hence

|f(x)− f(x0)| ≤ |f(x)− fN0(x)|+ |fN0(x)− fN0(x0)|+ |fN0(x0)− f(x0)|<

ε

3+ε

3+ε

3= ε if |x− x0| < δ.

Theorem 8.2.1′ Suppose that∞∑

n=1

un(x) is a series of continuous functions un(x) defined on the interval I

and converges uniformly to S(x). Then the sum S(x) is continuous on I .

8.2.2 Uniform Limit of Integrable Functions

Theorem 8.2.2 Suppose that {fn(x)} is a sequence of integrable functions on [a, b] and converges uniformly

to f(x) on [a, b]. Then the limit function f(x) is integrable, and moreover

limn→∞

∫ b

a

fn(x)dx =

∫ b

a

f(x)dx =

∫ b

a

limn→∞ fn(x)dx.

Proof. We first prove that f(x) is integrable in [a, b]. By the integrable criterion in Chapter 5, we need to

show that ∀ε > 0, we can find a partition P∗ on [a, b] such that

U(f, P∗)− L(f, P∗) < ε.

Since fn(x) converges uniformly to f(x) on [a, b], there is a sufficiently largeN 0 such that

fN0(x) −ε

3(b− a)≤ f(x) ≤ fN0(x) +

ε

3(b− a)∀x ∈ [a, b].


Hence for any partition P of [a, b],

L(fN0 , P )−ε

3≤ L(f, P ) ≤ U(f, P ) ≤ U(fN0, P ) +

ε

3.

Then we have

U(f, P )− L(f, P ) ≤ U(fN0 , P )− L(fN0 , P ) +2

3ε.

Since fN0 is integrable in [a, b], we can find a partition P∗ of [a, b] such that

U(fN0 , P∗)− L(fN0 , P∗) <1

3ε.

Therefore, for this partition P∗, we have

U(f, P∗)− L(f, P∗) <1

3ε+

2

3ε = ε.

Hence, f(x) is integrable on [a, b]. Next, we prove: ∀ε, ∃N such that

∣∣∣∣∣∫ b

a

fn(x)dx −∫ b

a

f(x)dx

∣∣∣∣∣ < ε, if n ≥ N.

Since {fn(x)} converges uniformly to f(x), there exists N such that for all x ∈ [a, b]

|fn(x)− f(x)| < ε

b− aif n ≥ N.

Hence ∣∣∣∣∣∫ b

a

fn −∫ b

a

f

∣∣∣∣∣ ≤∫ b

a

|fn − f | <∫ b

a

ε

b− a= ε, if n ≥ N,

which means that

limn→∞

∫ b

a

fn(x)dx =

∫ b

a

f(x)dx.


n=1

un(x) is a series of integrable functions un(x) on interval [a, b] and

converges uniformly to S(x). Then the sum S(x) is integrable on [a, b], and

limk→∞

∫ b

a

n∑k=1

uk(x)dx =

∫ b

a

S(x)dx,

or ∫ b

a

∞∑n=1

un(x)dx =

∞∑n=1

∫ b

a

un(x)dx.

In this case we say that the function series is integrable termwise.


8.2.3 Uniform Limit of Differentiable Functions

Theorem 8.2.3 Suppose that {fn(x)} is a sequence of continuously differentiable functions on the interval

[a, b] and satisfies

(i) The sequence {fn(x)} converges pointwise to the function f(x) on [a, b].

(ii) The sequence {f ′n(x)} converges uniformly to the function g(x) on [a, b].

Then the function f(x) is continuously differentiable and

f ′(x) = g(x), ∀x ∈ [a, b],

or

[ limn→∞ fn(x)]

′ = limn→∞ f ′

n(x).

Proof. According to the Fundamental Theorem of Calculus, for every n and all x ∈ [a, b]∫ x

a

f ′n(t)dt = fn(x)− fn(a).

From assumption (ii) and using Theorem 8.2.2, we have

limn→∞

∫ x

a

f ′n(t)dt =

∫ x

a

g(t)dt for a ≤ x ≤ b,

i.e.,

limn→∞[fn(x) − fn(a)] =

∫ x

a

g(t)dt.

By assumption (i),

f(x)− f(a) =

∫ x

a

g(t)dt, ∀x ∈ [a, b].

From assumption (ii) and using Theorem 8.2.1, we know that g(x) is continuous on [a, b]. Hence, by the

Intermediate Value Theorem of Integral, at each point x 0 ∈ [a, b]

1

Δx[f(x0 +Δx)− f(x0)] =

1

Δx

∫ x0+Δx

x0

g(t)dt

=1

Δxg(ξ) ·Δx,

where ξ is between x0 and x0 +Δx. As Δx→ 0, we have

f ′(x0) = g(x0), for x0 ∈ [a, b].



n=1

un(x) is a function series of which each un(x) is continuously differen-

tiable on the interval [a, b] and satisfies


(i)∞∑n=1

un(x) converges pointwisely to the function S(x) on [a, b]

(ii)∞∑n=1

u′n(x) converges uniformly to the function σ(x) on [a, b]

Then the function S(x) is continuously differentiable and

S′(x) = σ(x), ∀x ∈ [a, b],

or ( ∞∑n=1

un(x)

)′=

∞∑n=1

u′n(x).

In this case, we say that the function series is differentiable termwise.

8.3 Power Series

8.3.1 Radius of Convergence

Definition 8.3.1 A function series∞∑k=0

uk(x) is called a Power Series if its terms have the form

uk(x) = ak(x− x0)k, k = 0, 1, . . . ,

where ak are real numbers.

An important property of a power series is that it has a Radius of Convergence. In other words, the domain

of convergence of a power series∞∑k=0

ak(x − x0)k is an interval with the center x0 and a radius R, i.e.,

one of the following cases: [x0 − R, x0 + R], (x0 − R, x0 + R), [x0 − R, x0 + R), (x0 − R, x0 + R] or

(−∞,+∞). The last case can be understood as R = +∞. When R = 0, it means that the series converges

only at x0.

For∞∑k=0

ak(x−x0)k, a transform t = x−x0 will change it to∞∑k=0

aktk. Hence, we only need to discuss

the convergence of power series in the form∞∑k=0

akxk.

Lemma 8.3.2

(1) If∞∑k=0

akxk converges at the point x = x1 �= 0, then

∞∑k=0

akxk converges absolutely for |x| < |x1|,

i.e.∞∑k=0

|akxk| converges on the interval (−|x1|, |x1|)


2) If∞∑k=0

akxk does not converge at the point x = x2 �= 0, then

∞∑k=0

akxk does not converge for

|x| > |x2|, i.e. the series does not converge on (−∞,−|x2|) and (|x2|,∞).

Proof. From the convergence of the number series∞∑k=0

akxk1 , we know that lim

k→∞akx

k1 = 0. Hence, there

is a constant M > 0 such that

|akxk1 | ≤M, k = 0, 1, . . . , n, . . . .

For any x �= 0 satisfying |x| < |x1|

|akxk| = |akxk1 |∣∣∣∣ xx1∣∣∣∣k

≤M

∣∣∣∣ xx1∣∣∣∣k

.

Since

∣∣∣∣ xx1∣∣∣∣ < 1,

∞∑k=0

∣∣∣∣ xx1∣∣∣∣k

converges. By Comparison Test,∞∑k=0

|akxk| converges when |x| < |x1|. There-

fore,∞∑k=0

akxk converges absolutely in (−|x1|, |x1|).

We prove case (2) by contradiction. If there is a point x, which satisfies |x| > |x 2| and∞∑k=0

akxk

converges. Then from case (1),∞∑k=0

akxk2 converges absolutely. But it contradicts with the assumption, so

∞∑k=0

akxk does not converge.

Using Lemma 8.3.2 and the Nested Interval Theorem, we can prove the following theorem.

Theorem 8.3.3 For any power series∞∑k=0

akxk, there is a nonnegative number R (maybe 0 and +∞) such

that

(1) For |x| < R, the series converges absolutely.

(2) For |x| > R, the series diverges.

Proof. If the series converges everywhere, then R = +∞. If the series converges only at the point zero,

then R = 0. Now we need only to prove the case when the series converges at some point x 1 > 0 and

diverges at some point x2 > x1. In this case, we prove that there exists a number R > 0 such that the

series converges on (−R,R) and diverges on (−∞,−R) ∪ (R,+∞). Let u1 = x1, v1 = x2. Choose

c1 = (u1 + v1)/2. If the series converges at c1, let u2 = c1 and v2 = v1; otherwise let u2 = u1 and

v2 = c1. Step by step choose ck =uk + vk

2, and let uk+1 = ck and vk+1 = vk if the series converges

at ck; otherwise let uk+1 = uk and vk+1 = ck. Hence, we get two sequences {uk} and {vk}. The series∞∑k=0

akxk converges at all uk and diverges at all vk and {uk}. {vk} satisfy:

(1) {uk} and {vk} are monotonically increasing and decreasing respectively.


(2) limk→∞

(vk − uk) = limk→∞

x2 − x12k

= 0.

Therefore by the Nested Interval Theorem, there is a unique point R such that

uk ≤ R ≤ vk for all k

and

limk→∞

uk = limk→∞

vk = R.

Hence, for any x with |x| < R, there exists some index N such that |x| < uN . By the Lemma, the series∞∑k=0

akxk converges due to the convergence of the series

∞∑k=0

akukN . On the other hand, for any x with

|x| > R, there exists some index M such that |x| > vM . By the Lemma the series∞∑k=0

akxk diverges due

to the divergence of the series∞∑k=0

akvkM . Thus we complete the proof.

Note. The theorem does not give any conclusions on the two end points x = R and x = −R. The series

may converge or diverge at the end points. For example:∞∑k=1

xk converges on (−1, 1),∞∑k=1

xk

kconverges

on [−1, 1), and∞∑k=1

xk

k2converges on [−1, 1]. Three series have a same radius of convergence. But the

convergence of end points are different. They should be discussed individually.

Theorem 8.3.4 (Ratio Test) If the coefficients of a power series∞∑

n=0

anxn satisfy

limn→∞

|an+1||an| = l,

where l may be +∞. Then the radius of convergence is 1/l, or, we have

(1) if 0 < l < +∞, R = 1/l;

(2) if l = 0, R = +∞;

(3) if l = +∞, R = 0;

Proof. Let un(x) = anxn, and suppose x �= 0. Then∣∣∣∣un+1(x)

un(x)

∣∣∣∣ = |an+1||an| |x|.

By the assumption,

limn→∞

∣∣∣∣un+1(x)

un(x)

∣∣∣∣ = l |x|.

Hence,


(1) If 0 < l < +∞, by the Ratio Test for Series with Nonnegative Terms: l|x| < 1, then∞∑

n=0

|un(x)|

converges, i.e. |x| < 1/l, then∞∑n=0

un(x) converges absolutely. At the same time, if l|x| > 1, i.e.

|x| > 1/l, then {|un(x)|} doesn’t converge to zero, so {un(x)} doesn’t converge to zero also. Hence∞∑n=0

uk(x) doesn’t converge. Thus, we get the conclusionR = 1/l.

(2) If l = 0, by the assumption, for any x �= 0

limn→∞

|un+1(x)||un(x)| = 0.

Hence∞∑n=0

un(x) converges absolutely for any x �= 0. Since∞∑n=0

anxn converges obviously at x = 0,

hence∞∑

n=0

anxn converges on (−∞+∞), i.e., R = +∞.

(3) If l = +∞, that means for any x �= 0

limn→∞

|un+1(x)||un(x)| = lim

n→∞|an+1||an| |x| = +∞.

Then for n sufficiently large|un+1(x)||un(x)| > 1,

i.e., {un(x)} doesn’t converge to zero and∞∑

n=0

un(x) does not converge at any point x �= 0. Hence,

∞∑n=0

anxn converges only at x = 0, i.e., R = 0.

Theorem 8.3.5 (Root Test) If the coefficients of a power series∞∑

n=0

anxn satisfy

lim n√|an| = l,

where l may be +∞. Then the radius of convergence is 1/l.

8.3.2 Uniform Convergence

Theorem 8.3.6 Suppose that the radius of convergence of the power series∞∑n=0

anxn is R > 0. Then

(1) For any positive number b < R, the series∞∑n=0

anxn converges uniformly on [−b, b].

(2) If∞∑

n=0

anxn converges at x = R, then

∞∑n=0

anxn converges uniformly on [0, R].


(3) If∞∑

n=0

anxn converges at x = −R, then

∞∑n=0

anxn converges uniformly on [−R, 0].

Proof.

(1) We have proved that if |un(x)| ≤ cn on interval I and∞∑n=0

cn converges, then∞∑n=0

un(x) uniformly

converges in I . Now

|anxn| ≤ |an|bn on [−b, b]

and∞∑n=1

|an|bn converges. Hence∑anx

n converges uniformly on [−b, b]

(2),(3) Use Abel Test.

Theorem 8.3.7 If the radius of convergence of the power series∞∑n=0

anxn is R, then the series is continuous

on (−R,R). Moreover, if the series converges at x = R (or x = −R), then∞∑

n=0

anxn is also continuous at

x = R (or x = −R).

Proof. For any fixed point x0 ∈ (−R,R), there is a positive number b < R such that x0 ∈ (−b, b). From

Theorem 8.3.6, we know that∞∑n=0

anxn converges uniformly on [−b, b]. Obviously, each term anx

n of the

series is continuous on [−b, b], hence by Theorem 8.2.1’, the series is continuous in [−b, b], especially at

the point x0. Because x0 is any point in (−R,R),∞∑n=0

anxn is continuous in (−R,R)

If∞∑n=0

anxn also converges at x = R (or x = −R), from Theorem 8.3.6

∞∑n=0

anxn converges uniformly

on [0, R] (or [−R, 0]). Hence it is continuous on [0, R] (or [−R, 0]), especially continuous at x = R (or

x = −R).

Theorem 8.3.8 If the radius of convergence of power series∞∑

n=0

anxn is R, then

∫ x

0

∞∑n=0

antndt =

∞∑n=0

∫ x

0

antndt

=∞∑

n=0

ann+ 1

xn+1 for −R < x < R.

Proof. For x ∈ (−R,R),∞∑n=0

anxn converges uniformly on [−|x|, |x|], especially on [−|x|, 0] or [0, |x|].

Obviously, each term anxn is integrable. Hence by Theorem 8.2.2’, the series can be integrated termwisely.

Theorem 8.3.8′ If the radius of convergence of the power series∞∑n=0

anxn is R, and furthermore, if the


series converges at x = R, then

∫ R

0

∞∑n=0

antndt =

∞∑n=0

∫ R

0

antndt

=∞∑n=0

anRn+1

n+ 1

Proof. The conclusion comes from Theorem 8.3.6 directly.

Theorem 8.3.9 If the radius of convergence of the power series∞∑n=0

anxn is R, then

( ∞∑n=0

anxn

)′=

∞∑n=0

(anxn)′

=

∞∑n=0

nanxn−1, for −R < x < R

Proof. For any fixed point x ∈ (−R,R), there is a positive number b such that |x| < b < R. It is known

that∞∑n=0

anxn converges pointwise on [−b, b]. By the Theorem 8.2.3’, we need only to prove that the series

∞∑n=1

nanxn−1 converges uniformly on [−b, b]. In order to find a convergent series

∞∑n=1

cn which satisfies

cn > 0 and |nanxn−1| ≤ cn in [−b, b], we choose a number r such that b < r < R. Then for x ∈ [−b, b]

|nanxn−1| ≤ n|an|bn−1 =n

r|anrn|

(b

r

)n−1

.

Since∞∑n=1

|anrn| converges, {|anrn|} is bounded, i.e., there exists a constant M > 0 such that

|anrn| ≤M, n = 1, 2, . . . .

Hence

|nanxn−1| ≤ M

rn

(b

r

)n−1

= Cn.

Due tob

r< 1, the series

∞∑n=1

cn converges. Hence∞∑n=1

nanxn−1 converges uniformly in [−b, b]. By the

Theorem 8.2.3’, the series can be differentiated termwisely on [−b, b], especially at the point x which we

choose. Since x can be any point in (−R,R), the formula holds on (−R,R).In the above proof, we can see that the convergent radius of

∞∑n=1

nanxn−1 is also the same as

∞∑n=1

anxn.

Iteratively apply Theorem 8.3.9, we get the following corollary.


Corollary 8.3.10 Suppose that the radius of convergence of a power series is R. Then the power series can

be differentiated termwisely for k times, k is any natural number. That is

S(k)(x) =

∞∑n=0

(anxn)(k) =

∞∑n=k

n(n− 1) · · · (n− k + 1)xn−k k = 1, 2, . . . .

The convergent radius of S (k)(x) is also the same as S(x).

Example 8.3.1 Find the domain of convergence of the following power series:

(a)∞∑

n=1

xn

n2n.

(b)∞∑

n=1

n!

nnxn.

(c)∞∑n=1

xn√n

.

(d)∞∑n=1

n!

nn(x− 2)n.

8.4 Taylor Series

8.4.1 Taylor Polynomials and Order of Contact

Definition 8.4.1 Let functions f(x) and g(x) be defined on an open interval I and x 0 ∈ I . We say that

f(x) and g(x) have contact of order n at x0 provided that

f (k)(x0) = g(k)(x0) for k = 0, 1, . . . , n.

Lemma 8.4.2 Let I be an open interval and x0 ∈ I . Suppose that the function f(x) has (n+1) derivatives

on I , and

f (k)(x0) = 0 for k = 0, . . . , n.

Then for each point x ∈ I and x �= x0, there is a point ξ strictly between x and x0 at which

f(x) =f (n+1)(ξ)

(n+ 1)!(x− x0)

n+1.

Corollary 8.4.3 If f(x) and g(x) have contact of order n at x0, then

f(x)− g(x) =f (n+1)(ξ)− g(n+1)(ξ)

(n+ 1)!(x − x0)

n+1x0 < ξ < x or x < ξ < x0.

Corollary 8.4.3 shows that f(x) and g(x) are very closed to each order in the neighborhood of x 0.

Polynomials are the simplest kind of functions. Hence, for a general function f(x) it is natural to seek

a polynomial that is a good approximation of f(x).


Proposition 8.4.4 Suppose that the function f(x) has n derivatives on an open interval I . Then there is a

unique polynomial of degree at most n that has contact of order n with the function f at point x 0 ∈ I . This

polynomial is defined by the formula

pn(x) = f(x0) + f ′(x0)(x − x0) + · · ·+ f (n)(x0)

n!(x− x0)

n

=

n∑k=0

f (k)(x0)

k!(x − x0)

k.

Proof. Let pn(x) =n∑

k=0

ck(x − x0)k be a polynomial that has contact of order n with function f(x). By

the definition,

p(k)n (x0) = f (k)(x0), k = 0, 1, . . . , n.

But p(k)n (x0) = k!ck. Hence, ck = f(k)(x0)k! .

Definition 8.4.5 For the function f(x), the polynomial

pn(x) =

n∑k=0

f (k)(x0)

k!(x− x0)

k

is called the nth Taylor polynomial at the point x0

Theorem 8.4.6 (Lagrange Remainder Theorem) Suppose that the function f(x) has (n+ 1) derivatives

on the open interval I and pn(x) is the nth Taylor polynomial for f(x) at the point x0 ∈ I , i.e.

pn(x) =

n∑k=0

f (k)(x0)

k!(x− x0)

n.

Then for each point x ∈ I , there is a point ξ strictly between x and x0 such that

f(x)− pn(x) =f (n+1)(ξ)

(n+ 1)!(x− x0)

n+1.

Proof. Since f(x) and pn(x) have contact of order n, from Lemma 8.4.2 we obtain

f(x) − pn(x) =f (n+1)(ξ)− p

(n+1)n (ξ)

(n+ 1)!(x− x0)

n+1.

But p(n+1)n (x) ≡ 0, it follows that

f(x)− pn(x) =f (n+1)(ξ)

(n+ 1)!(x− x0)

n+1.

The difference between f(x) and pn(x) is called the Remainder of the Taylor Polynomial approximation.

Definition 8.4.7 Let I be a neighborhood of x0. If f : I → R has derivatives of all orders. Let

pn(x) =

n∑k=0

f (k)(x0)

k!(x− x0)

k.


If

limn→∞ pn(x) = f(x) ∀x ∈ I,

we write

f(x) =

∞∑k=0

f (k)(x0)

k!(x− x0)

k ∀x ∈ I.

This formula is called a Taylor Series expansion of f(x) about x0. If x0 = 0, then the series

∞∑n=0

f (n)(0)

n!xn ∀x ∈ I

is called Maclurin Series of f(x).

Theorem 8.4.8 Suppose that the power series∞∑

n=0

an(x − x0)n converges to f(x) on (x0 − R, x0 + R).

Then the series must be the Taylor Series of f(x) expanding at x = x0. That is

an =1

n!f (n)(x0), n = 0, 1, 2, . . . .

Note: Sometime it is difficult to determine the domain of convergence analyzing the Remainder Term of the

Taylor Polynomial. Using this theorem, we may get the convergence by applying the theory of power series.

Theorem 8.4.9 (Cauchy Integral Remainder Formula) Suppose that the function f(x) has n+1 deriva-

tives on an open interval I and f (n+1)(x) is continuous on I . Then for each point x in I ,

f(x)−n∑

k=0

f (k)(x0)

k!(x− x0)

k =1

n!

∫ x

x0

f (n+1)(t)(x− t)ndt.

Proof. By the First Fundamental Theorem of Calculus,

f(x)− f(x0) =

∫ x

x0

f ′(t)dt.

Integrating by parts, ∫ x

x0

f ′(t)dt = −∫ x

x0

f ′(t)d(x − t)

= −f ′(t)(x − t)

∣∣∣∣x

x0

+

∫ x

x0

f ′′(t)(x − t)dt

= f ′(x0)(x − x0) +

∫ x

x0

f ′′(t)(x − t)dt,

which is the result for n = 1. The general formula follows by induction. If the theorem is true for m, i.e.

f(x) =

m∑k=0

f (k)(x0)

k!(x− x0)

k +1

m!

∫ x

x0

f (m+1)(t)(x − t)mdt,


then

1

m!

∫ x

x0

f (m+1)(t)(x − t)mdt =−1

(m+ 1)!

∫ x

x0

f (m+1)(t)d(x − t)m+1

=−1

(m+ 1)!f (m+1)(t)(x − t)m+1

∣∣∣∣x

x0

+1

(m+ 1)!

∫ x

x0

f (m+2)(t)(x − t)m+1dt

=1

(m+ 1)!f (m+1)(x0)(x − x0)

m+1 +1

(m+ 1)!

∫ x

x0

f (m+2)(t)(x− t)m+1dt

We obtain

f(x) =m+1∑k=0

f (k)(x0)

k!(x − x0)

k +1

(m+ 1)!

∫ x

x0

f (m+2)(t)(x− t)m+1dt

That means the theorem is also true for m+ 1.

8.4.2 Convergence of Taylor Series

Example 8.4.1 Let a > 0 and

f(x) =

⎧⎨⎩ e−

1x2 , x �= 0,

0, x = 0.

Show that the series∞∑n=0

f (n)(0)

n!xn converges on (−a, a), but does not converge to f(x).

Proof. It is easy to see that f (n)(x) exists on (−a, a) for any n, and f (n)(0) = 0, ∀n. Then, we have

∞∑n=0

f (n)(0)

n!xn = 0 �= f(x), ∀x ∈ (−a, a), except x = 0.

Lemma 8.4.9 For any number c

limn→∞

cn

n!= 0.

Proof. Choose k to be a natural number such that k ≥ 2|c|. Then for n ≥ k

0 ≤∣∣∣∣cnn!∣∣∣∣

=

( |c|1

· · · |c|k

)( |c|k + 1

· · · |c|n

)

≤ |c|k(1

2

)n−k

= (2|c|)k(1

2

)n

.

2|c|k is a constant and limn→∞

(1

2

)n

= 0. Hence by squeezing principle, lim

∣∣∣∣cnn!∣∣∣∣ = 0 and so lim

n→∞cn

n!= 0.


Theorem 8.4.10 Suppose that f(x) has derivatives of all order on the interval [x 0 − r, x0 + r] and

|f (n)(x)| ≤Mn for x0 − r ≤ x ≤ x0 + r,

where r andM are positive and n is any natural number. Then

∞∑k=0

fk(x0)

k!(x− x0)

k = f(x) ∀x ∈ [x0 − r, x0 + r].

Proof. According to the Lagrange Remainder Theorem, for each x ∈ [x 0 − r, x0 + r]∣∣∣∣∣f(x)−n∑

k=0

f (k)(x0)

k!

∣∣∣∣∣ = |f (n+1)(ξ)|(n+ 1)!

|x− x0|n+1,

where ξ is strictly between x and x0. Hence,∣∣∣∣∣f(x)−n∑

k=0

f (k)(x0)

k!(x− x0)

k

∣∣∣∣∣ ≤ Mn+1

(n+ 1)!|x− x0|n+1 ≤ cn+1

(n+ 1)!,

where C =Mr. The conclusion follows immediately from Lemma 8.4.9.

Example 8.4.2 Let f(x) = ex. Since

f (n)(x) = ex, f (n)(0) = 1, n = 0, 1, 2, · · · .

Then for any r > 0 and for any n ≥ 0 we have

|f (n)(x)| = ex ≤ er ∀x ∈ [−r, r].

Therefore,

ex =

∞∑k=0

xk

k!∀x ∈ (−∞,+∞).

Example 8.4.3 Let f(x) = sinx. Since

f (n)(x) = sin(x+

nπ

2

), f (n)(0) = sin

nπ

2=

⎧⎨⎩ 0, n even,

(−1)(n−1)/2, n odd.

Then for any n ≥ 0 we have

|f (n)(x)| =∣∣∣ sin(x+

nπ

2

)∣∣∣ ≤ 1 ∀x ∈ (−∞,+∞).

Therefore,

sinx =

∞∑k=0

(−1)k

(2k + 1)!x2k+1 ∀x ∈ (−∞,+∞).

Example 8.4.4 Similarly, we have

cosx =

∞∑k=0

(−1)k

(2k)!x2k ∀x ∈ (−∞,+∞).


Example 8.4.5 Let f(x) = ln(1 + x), x > −1. Since

f (n)(x) =(−1)n−1(n− 1)!

(1 + x)n, f (n)(0) = (−1)n+1(n− 1)!, n ≥ 1.

Then we have the series∞∑k=1

(−1)k−1

kxk. (8.4.1)

We need to determine the domain of convergence of the series. Consider the series

∞∑k=0

(−1)kxk =1

1 + xx ∈ (−1, 1).

The radius of convergence of this power series is R = 1, then from Theorem 8.3.8 we have

ln(1 + x) =

∫ x

0

dx

1 + x

=

∫ x

0

∞∑k=0

(−1)kxk

=

∞∑k=0

∫ x

0

(−1)kxk

=

∞∑k=1

(−1)k−1

kxk, x ∈ (−1, 1).

Moreover, since the series (1) converges at x = 1, from Theorem 8.3.7

ln(1 + x) =

∞∑k=1

(−1)k−1

kxk x ∈ (−1, 1].

Example 8.4.6 Prove that∞∑

n=1

(−1)n−1

n= ln 2.

Example 8.4.7 (Newton’s Binomial Expansion) Let f(x) = (1 + x)α, where α is a real number. Since

f (n)(x) = α(α − 1) · · · (α− n+ 1)(1 + x)α−n, f (n)(0) = α(α− 1) · · · (α− n+ 1).

Then we have the series

1 +∞∑k=1

α(α− 1) · · · (α− k + 1)

k!xk, −1 < x < 1.

Proof. Using Cauchy Integral Remainder we have

Rn(x) = f(x)−n∑

k=0

f (k)(x0)

k!(x− x0)

k

=1

n!

∫ x

x0

f (n+1)(t)(x− t)ndt

=1

n!f (n+1)(ξ)(x − ξ)n(x− x0),


where ξ is between x0 and x. Let ξ = x0 + θ(x − x0), 0 < θ < 1, then

Rn(x) =1

n!f (n+1)(x0 + θ(x− x0))(1 − θ)n(x − x0)

n+1.

For x0 = 0 we have

Rn(x) =1

n!f (n+1)(θx)(1 − θ)nxn+1

=1

n!α(α− 1) · · · (α− n)(1 + θx)α−n−1(1− θ)nxn+1

=1

n!α(α− 1) · · · (α− n)xn+1(1 + θx)α−1

( 1− θ

1 + θx

)n.

Consider the series∞∑k=0

α(α− 1) · · · (α− k)

k!xk+1.

It is easy to see that the radius of convergence of this series is R = 1. Therefore, for |x| < 1

limn→+∞

α(α − 1) · · · (α− n)

n!xn+1 = 0.

Since 0 < 1− |x| ≤ |1 + θx| ≤ 1 + |x| < 2, let M(x) = max{(1− |x|)α−1, 2α−1}, then

|1 + θx|α−1 ≤ M(x), i.e., for fixed x, |1 + θx|α−1 is bounded. Also, for |x| < 1, 0 < 1 − θ < 1 + θx,

then 0 <( 1− θ

1 + θx

)n< 1. Therefore,Rn(x) → 0, ∀|x| < 1.

Example 8.4.8 Let f(x) = sin−1 x. It’s somewhat complicated to expand by directly differentiating f(x)

times by times. It’s known that

sin−1 x =

∫ x

0

dt√1− t2

for x ∈ (−1, 1).

By Newton’s Binomial Expansion, we know

1√1− t2

=

∞∑k=0

⎛⎝ 1

2

k

⎞⎠ (−t2)k =

∞∑k=0

(2k − 1)!!

2k!!x2k.

By Applying Theorem 8.3.8 (Termwise integrate the uniformly convergent series), for any x ∈ (−1, 1)

sin−1 x =

∫ x

0

dt√1− t2

=

∞∑k=0

(2k − 1)!!

2k!!

x2k+1

2k + 1.

Example 8.4.9 Let f(x) = tan−1 x. Expand1

1 + x2to a power series

1

1 + x2= 1− x2 + x4 − x6 + · · ·

=

∞∑n=0

(−1)nx2n


The convergent radius of the series is R = 1. By Theorem 8.3.8, we have∫ x

0

1

1 + t2dt =

∞∑n=0

∫ x

0

(−1)nt2ndt − 1 < x < 1

⇒ tan−1 x =

∞∑n=0

(−1)nx2n+1

2n+ 1, −1 < x < 1

The series is convergent at the end points x = 1 as well as x = −1.

Note: The derivative of the series∞∑

n=0

(−1)nx2n+1

2n+ 1is

∞∑n=0

(−1)nx2n which is not convergent at both end

points -1 and 1. This example shows that although the convergent radius R will not change for differentia-

tion of a power series, the convergence at the end points will not preserve.

8.5 Exercise

1. For each natural number n and each number x, define

fn(x) =1− |x|n1 + |x|n .

Find the function f : R → R to which the sequence {fn : R → R} converges pointwise. Prove

that the convergence is not uniform.

Hint. fn continuous, fn(0) = 1 and fn(1) = 0. By Intermediate Value Theorem, ∀n, ∃xn ∈ (0, 1), s.t.fn(xn) = 1/2. Since f(xn) = 1, then, |f(xn)− fn(xn)| = 1/2.

2. For each natural number n and each number x ≥ 2, define

fn(x) =1

1 + xn.

Find the function f : [2,∞) → R to which the sequence {fn : [2,∞) → R} converges pointwise.

Prove that the convergence is uniform.

3. For each natural number n and each number x ∈ (0, 1), define

fn(x) =1

nx+ 1.

Find the function f : (0, 1) → R to which the sequence {fn : (0, 1) → R} converges pointwise.

Prove that the convergence is not uniform.

Hint. ∀n, let xn = 1/n, then fn(xn) = 1/2, and |f(xn)− fn(xn)| = fn(xn) = 1/2.

4. For each natural number n and each number x ∈ [0, 1], define

fn(x) =x

nx+ 1.

Find the function f : [0, 1] → R to which the sequence {fn : [0, 1] → R} converges pointwise.

Prove that the convergence is uniform.

Hint. |f(x)− fn(x)| = |0− x/(nx+ 1)| = 1/(n+ 1/x) < 1/n if x �= 0, and |f(0) − fn(0)| = 0.


5. For each natural number n and each number x ∈ (−1, 1), define

pn(x) = x+ x(1− x2) + · · ·+ x(1 − x2)n.

Prove that the sequence {pn : (−1, 1) → R} converges pointwise.

6. Suppose that the sequence {fn : D → R} and {gn : D → R} converge uniformly to the functions

f : D → R and g : D → R respectively. For any two numbers α and β, prove that the sequence

{αfn + βgn : D → R} converges uniformly to the function αf + βg : D → R.

7. For each natural number n, let the function fn : R → R be bounded. Suppose that the sequence

{fn} converges uniformly to f on R. Prove that the limit function f : R → R is also bounded.

8. Let {an} be a bounded sequence of numbers. For each natural number n and each number x, define

fn(x) = a0 + a1x+a2x

2

2!+ · · ·+ anx

n

n!.

Prove that for each r > 0, the sequence of functions {fn : [−r, r] → R} is uniformly convergent.

9. Let fn(x) =√x2 + 1/n and f(x) = |x|. Prove that the sequence {fn} converges uniformly on the

open interval (−1, 1) to the function f . Check that each function fn is continuously differentiable,

whereas the limit function f is not differentiable at x = 0. Does this contradict Theorem 8.2.3?

10. Let fn(x) = nxe−nx2

. Prove that {fn(x)} converges pointwise to f(x) = 0 on the interval [0, 1],

but that the sequence{∫ 1

0

fn(x)dx}

does not converge to 0. Does this contradict Theorem 8.2.2?

11. Let fn(x) = nxe−nx2

. Prove that {fn(x)} converges uniformly to f(x) = 0 on the interval

[δ, 1], 0 < δ < 1, but does not converge uniformly to f(x) = 0 on the interval [0, 1].

12. Prove that if {fn : R → R} is a sequence of continuously differentiable functions such that the

sequence of derivatives {f ′n : R → R} is uniformly convergent and the sequence {fn(0)} is also

convergent, then {fn : R → R} is pointwise convergent. Is the assumption that the sequence

converges necessary? (Consider fn(x) = n ∀x ∈ R.)

13. Give an example of a sequence of differentiable functions f n : (−1, 1) → R that converges uni-

formly but for which {f ′n(0)} is unbounded. (fn(x) = (sinnx)/

√n.)

14. Let {an} and {bn} be bounded sequences. Prove that the series

∞∑n=1

1

n2(an cosnx+ bn sinnx)

converges uniformly on (−∞,+∞).


15. Determine the domain for conditional convergence and absolute convergence of the following series:

(a)∞∑

n=1

(−1)n+1xn

n

(b)∞∑

n=1

(−1)n

2n− 1

(1− x

1 + x

)n(c)

∞∑n=1

xn

n!

(d)∞∑n=1

xn

1− xn

16. Prove that the series ∞∑n=1

(−1)n

n(xn + 1)

converges uniformly on [1,+∞). Hint. Use Abel Test.

17. Prove that the series ∞∑n=1

(−1)n

x+ n

converges uniformly on (0,+∞). Hint. Use Dirichlet Test.

18. Determine the domain of convergence of each of the following power series:

(a)∞∑k=1

xk

k5k

(b)∞∑k=1

k!xk

(c)∞∑k=0

(−1)kx2k−1

(2k + 1)!


(a)1

1− x=

∞∑k=0

xk, |x| < 1.

(b)1

1 + x=

∞∑k=0

(−1)kxk, |x| < 1.

(c)1

(1 + x)2=

∞∑k=1

(−1)k+1kxk−1, |x| < 1.

20. Use (b) in the above problem to obtain a series expansion for the integral∫ 1/2

0

1

1 + x4dx and justify

your calculation.

21. Prove that1

(1 + x2)2=

∞∑k=0

(−1)k+1kx2k−2, |x| < 1.

22. Prove that x =

∞∑k=0

(1− 1

x

)k, |1− x| < |x|.

23. Let f(x) =1

(1− x)3, |x| < 1. Find a power series expansion for the function f : (−1, 1) → R.

24. Suppose that the domain of convergence of the power series∞∑k=0

ckxk contains the interval (−r, r).

Define f(x) =∞∑k=0

ckxk, |x| < r. Let the interval [a, b] be contained in the interval (−r, r). Prove

that ∫ b

a

f(x)dx =∞∑k=0

ckk + 1

(bk+1 − ak+1).

Chapter 9: Fourier Series 179

25. Find the Taylor polynomial

p3(x) =

3∑k=0

f (k)(1)

k!(x − 1)k

and the Lagrange Remainder for the function f(x) = 1 + 3x+ 5x2 − 2x3.

26. Find the Taylor series expansion about 0 for the function f(x) = (1− x)/(1 + x).

27. Suppose that f(x) is continuous on (−∞,∞). Let

fn(x) =

n−1∑k=0

1

nf(x+ k/n).

Prove that the function sequence {fn(x)} converges uniformly on any finite interval.

Hint. Since limn→+∞ fn(x) = lim

n→+∞

n−1∑k=0

1

nf(x+ k/n) =

∫ 1

0f(x+ t)dt, and f(x) is continuous, then {fn(x)}

converges for x ∈ (−∞,+∞). Rewrite fn(x) and its limit as

fn(x) =

n−1∑k=0

∫ (k+1)/n

k/nf(x+ k/n)dt,

∫ 1

0f(x+ t)dt =

n−1∑k=0

∫ (k+1)/n

k/nf(x+ t)dt.

Then we have∣∣∣fn(x)−∫ 1

0f(x+ t)dt

∣∣∣ =∣∣∣n−1∑k=0

∫ (k+1)/n

k/n[f(x+ k/n)− f(x+ t)]dt

∣∣∣

≤n−1∑k=0

∫ (k+1)/n

k/n|f(x+ k/n)− f(x+ t)|dt.

Since f(x) is uniformly continuous on any finite interval, then for given ε > 0, ∃δ > 0, s.t., |f(x1)− f(x2)| < ε,

∀x1, x2 ∈ [a − 1, b + 1], |x1 − x2| < δ. For x ∈ [a, b], take N s.t.,1

N< δ, then for t ∈ [ k

N,k + 1

N

],

∣∣x+k

n− (x+ t)

∣∣ < δ, and |f(x+ k/n)− f(x+ t)| < ε, ∀n ≥ N . Therefore,∣∣∣fn(x)−

∫ 1

0f(x+ t)dt

∣∣∣ < ε,

∀x ∈ [a, b],∀n ≥ N , i.e., {fn(x)} converges uniformly on any finite interval.


Chapter 9

Fourier Series

9.1 Fourier Series and Fourier Coefficients

In physics and engineering, periodic phenomenon are frequently encountered. To describe these phe-

nomenon, Periodic Functions are required. The general periodic functions are defined as

f(t+ T ) = f(t).

The number T is called the period. The well known periodic functions are Trigonometric functions sinx

and cosx which have the period T = 2π. Now, we try to expand any periodic function f(x) with period T

to a series of sin and cos function. That is

f(x) =a02

+∞∑k=1

(ak cos kωx+ bk sin kωx),

where ω = 2π/T . For simplicity, we consider periodic functions f(x) with period 2π and then we have

f(x) =a02

+

∞∑k=1

(ak cos kx+ bk sin kx).

Definition 9.1.1 A function set {φk(x)} is said to be orthogonal on [a, b] provided

∫ b

a

φm(x)φn(x)

⎧⎨⎩ = 0, if m �= n,

> 0, if m = n.

Theorem 9.1.2 The function set {1, cosx, sinx, cos 2x, sin 2x, · · · , cosnx, sinnx, · · · } is orthogonal

on [−π, π].According to the theory of function series, if

f(x) =a02

+

∞∑k=1

(ak cos kx+ bk sin kx)

181


converges uniformly on [−π, π], then multiplying cosnx on both sides and integrating from −π to π, we

get

an =1

π

∫ π

−π

f(x) cosnx, n = 0, 1, 2, · · · .

Multiplying sinnx on both sides and integrating from −π to π, we get

bn =1

π

∫ π

−π

f(x) sinnx, n = 1, 2, · · · .

In general, the assumption of ”uniform convergence” is not reasonable. In fact we often want to use

such kind of trigonometric series to approximate a discontinuous function f(x). By the theory of function

series, because trigonometric functions are continuous, the sum of the Trigonometric series must be con-

tinuous if the series is uniformly convergent. Hence the trigonometric series can’t converge uniformly to a

discontinuous function.

Definition 9.1.3 For a periodic function f(x) with period 2π defined on (−∞,+∞), the trigonometric

series

a02

+

∞∑n=1

(an cosnx+ bn sinnx)

is called the Fourier Series of f(x), provided that the coefficients an, bn are calculated by the above formula.

We denote this by

f(x) ∼ a02

+

∞∑n=1

(an cosnx+ bn sinnx).

The coefficients an, bn are called the Fourier Coefficients of f(x).

Note 1: The symbol ”∼” doesn’t mean equal, since we still do not know whether the series converges and

whether it converges to f(x).

Note 2: If the period of f(x) is T , then

f(x) ∼ a02

+∞∑n=1

(an cosnωx+ bn sinnωx),

where ω = 2π/T and

an =2

T

∫ T/2

−T/2

f(x) cosnωx, n = 0, 1, . . .

bn =2

T

∫ T/2

−T/2

f(x) sinnωx, n = 1, . . .


9.2 Convergence of Fourier Series

Theorem 9.2.1 (Weierstrass Approximation Theorem) Suppose that f(x) is continuous on [a, b], then for

given ε > 0, there exists a polynomial pn(x),

pn(x) =

n∑k=0

f(kn

)( k

n

)xk(1− x)n−k,

such that

|f(x)− pn(x)| < ε, ∀x ∈ [a, b].

Proof. We only need to prove the case [a, b] = [0, 1]. Assume that |f(x)| ≤ M . Since f(x) is uniformly

continuous, then for given ε > 0, there exists a δ > 0, such that

|f(x)− f(x′)| < ε/2, ∀x ∈ [0, 1].

Since

f(x) =n∑

k=0

f(x)( k

n

)xk(1− x)n−k,

then we have

|f(x)− pn(x)| ≤n∑

k=0

∣∣∣f(x)− f(kn

)∣∣∣( k

n

)xk(1− x)n−k.

Divide k = 0, 1, · · · , n into two sets:

k ∈ A∣∣∣kn− x∣∣∣ < δ,

k ∈ B∣∣∣kn− x∣∣∣ ≥ δ.

If k ∈ A, we have

∑k∈A

∣∣∣f(x)− f(kn

)∣∣∣( k

n

)xk(1− x)n−k <

ε

2.

If k ∈ B, using the following lemma we get

∑k∈B

∣∣∣f(x)− f(kn

)∣∣∣( k

n

)xk(1 − x)n−k ≤ 2M

n2δ2

∑k∈B

(k − nx)2( k

n

)xk(1− x)n−k

≤ M

2nδ2<ε

2.


Lemma 9.2.2

n∑k=0

(k − nx)2( k

n

)xk(1− x)n−k ≤ n

4, ∀x ∈ [0, 1].

Note 1: Weierstrass Approximation Theorem is also called Bernstein Approximation Theorem.

Using Weierstrass Approximation Theorem, we get

Lemma 9.2.3 (Riemann’s Lemma) Suppose that f(x) is integrable on [a, b]. Then we have

limλ→+∞

∫ b

a

f(x) sinλxdx = 0, limλ→+∞

∫ b

a

f(x) cosλxdx = 0.

Proof. The proof is divided into the following four steps:

(i) For given ε > 0, there exists a continuous function g(x) such that∫ b

a

|f(x)− g(x)|dx < ε

3.

(ii) For given ε > 0, there exists a polynomial p(x) such that

|g(x)− p(x)|dx < ε

3(b− a), ∀x ∈ [a, b].

(iii)

limλ→+∞

∫ b

a

p(x) sinλxdx = 0.

(iv) ∣∣∣ ∫ b

a

f(x) sin λxdx∣∣∣ ≤

∣∣∣ ∫ b

a

[f(x)− g(x)] sin λxdx∣∣∣ +

∣∣∣ ∫ b

a

[g(x)− p(x)] sin λxdx∣∣∣ + ∣∣∣ ∫ b

a

p(x) sin λxdx∣∣∣

≤∫ b

a

|f(x)− g(x)|dx +

∫ b

a

|g(x)− p(x)|dx +ε

3< ε.

Suppose that the period function f(x) is integrable on [−π, π], we consider the convergence of the

Fourier series of f(x). For any given point x, we have

Sn(x) =a02

+

n∑k=1

(ak cos kx+ bk sin kx)

=1

π

∫ π

−π

f(t)[12+

n∑k=1

(cos kt cos kx+ sin kt sin kx)]dt

=1

π

∫ π

−π

f(t)[12+

n∑k=1

(cos k(t− x)]dt

=1

π

∫ π

−π

f(t)sin[(2n+ 1)(t− x)/2]

2 sin[(t− x)/2]dt


Let u = t− x, we get

Sn(x) =1

π

∫ π−x

−π−x

f(u+ x)sin[(2n+ 1)u/2]

2 sin(u/2)du

=1

π

∫ π

−π

f(u+ x)sin[(2n+ 1)u/2]

2 sin(u/2)du

=1

π

∫ π

0

[f(u+ x) + f(x− u)]sin[(2n+ 1)u/2]

2 sin(u/2)du.

If we take f(x) ≡ 1, then we get

1

π

∫ π

0

sin[(2n+ 1)u/2]

sin(u/2)du = 1

Let S = [f(x+ 0) + f(x− 0)]/2, then we have

Sn(x)− S =1

π

∫ π

0

φ(u)sin[(2n+ 1)u/2]

2 sin(u/2)du,

where

φ(u) = f(x+ u) + f(x− u)− f(x+ 0)− f(x− 0).

Thus, we have the following lemma:

Lemma 9.2.4 Suppose that f(x) is periodic with period 2π and integrable on [−π, π]. Also, suppose that

at the point x the left and right limits f(x− 0) and f(x+0) exist. Then Sn(x) converges to S if and only if

limn→∞

∫ π

0

φ(u)sin[(2n+ 1)u/2]

sin(u/2)du = 0.

Since the function

φ(u)[ 1

sin(u/2)− 1

(u/2)

]is integrable on [0, π], then from Riemann’s Lemma we get

limn→∞

∫ π

0

φ(u)[ 1

sin(u/2)− 1

(u/2)

]sin[(2n+ 1)u/2]du = 0.

Therefore, Lemma 9.2.4 can be rewritten as

Lemma 9.2.5 Suppose that f(x) is periodic with period 2π and integrable on [−π, π]. Also, suppose that

at the point x the left and right limits f(x− 0) and f(x+0) exist. Then Sn(x) converges to S if and only if

limn→∞

∫ π

0

φ(u)sin[(2n+ 1)u/2]

udu = 0.

For any given δ > 0, the function φ(u)/u is integrable on [δ, π]. Then from Riemann Lemma

limn→∞

∫ π

δ

φ(u)sin[(2n+ 1)u/2]

udu = 0.


Using this we have

Theorem 9.2.6 (Riemann’s Localization Theorem) Suppose that f(x) is periodic with period 2π and

integrable on [−π, π]. Also, suppose that at the point x the left and right limits f(x− 0) and f(x+0) exist.

Then Sn(x) converges to S if there exists a δ > 0 such that

limn→∞

∫ δ

0

φ(u)sin[(2n+ 1)u/2]

udu = 0,

i.e., the convergence of the Fourier series of f(x) at the point x depends only on the function values of f(x)

in the neighborhood of x.

Theorem 9.2.7 (Lipschitz Test) Suppose that f(x) is periodic with period 2π and integrable on [−π, π].Also, suppose that at the point x the left and right limits f(x − 0) and f(x + 0) exist, and there exist a

constant L and δ0 such that

|f(x± u)− f(x± 0)| ≤ Lu, 0 < u < δ.

Then Sn(x) converges to S.

Proof.φ(u)

u=f(x+ u)− f(x+ 0)

u+f(x− u)− f(x− 0)

u.

Consider

g(u) =f(x+ u)− f(x+ 0)

u.

By assumption, g(u) is bounded on [0, δ0], i.e., |g(u)| ≤ L. Since f(x) is integrable, then for any η > 0,

g(u) is integrable on [η, δ0], and g(u) is integrable on [0, δ0]. Similarly,f(x− u)− f(x− 0)

uis integrable

on [0, δ0]. Therefore,φ(u)

uis integrable on [0, δ0]. From Riemann’s Lemma,

limn→∞

∫ δ

0

φ(u)sin[(2n+ 1)u/2]

udu = 0.

From Theorem 9.2.6, Sn(x) converges to S.

Corollary 9.2.8 Suppose that f(x) is periodic with period 2π and integrable on [−π, π]. Also, suppose that

at the point x, f(x) is differentiable, or the left and right derivatives of f(x) exist. Then Sn(x) converges

to f(x).

Proof. Suppose

f ′+(x) = lim

u→0+

f(x+ u)− f(x+ 0)

u= A,

then, ∃δ1 > 0, s.t., ∣∣∣f(x+ u)− f(x+ 0)

u−A∣∣∣ ≤ 1,


or,

|f(x+ u)− f(x+ 0)| ≤ (|A|+ 1)u = L1u, 0 < u < δ1.

Similarly,

|f(x− u)− f(x− 0)| ≤ L2u, 0 < u < δ2.

Let L = max{L1, L2}, then

|f(x± u)− f(x± 0)| ≤ L|u|, 0 < u < δ0 = min{δ1, δ2}.

From Theorem 9.2.7, Sn(x) converges to f(x).

Definition 9.2.9 A function f(x) is called piecewise continuous on [a, b], if there exist finite number of points

a = x0 < x1 < · · · < xn = b such that f(x) is continuous on every subinterval (xi−1, xi), i = 1, 2, · · · , n,

and the limits f(a+0), f(xi − 0), f(xi +0), i = 1, 2 · · · , n− 1, and f(b− 0) exist. If f(x) and f ′(x) are

piecewise continuous on [a, b], we say that f(x) is piecewise smooth on [a, b].

Lemma 9.2.10 Suppose that f(x) is piecewise smooth on [a, b], then

limx→c+

f(x)− f(c+ 0)

x− c= f ′(c+ 0), if c ∈ [a, b)

and

limx→c−

f(x)− f(c− 0)

x− c= f ′(c− 0), if c ∈ (a, b].

Proof. If c �= xi, i = 0, 1, · · · , n, the lemma is obviously true. If c = xi, i = 0, 1, · · · , n, redefine

f(xi) = f(xi + 0), then f(x) is continuous on [xi, x], for x ∈ (xi, xi+1). Using Lagrange Mean Value

Theorem, there exists ξ ∈ (xi, x) such that,

f(x)− f(xi + 0)

x− xi= f ′(ξ),

and

limx→xi+

f(x)− f(xi + 0)

x− xi= f ′(xi + 0).

Theorem 9.2.11 Suppose that f(x) is periodic with period 2π and piecewise smooth on [−π, π]. Then

Sn(x) converges to S.

Proof. Using Lemma 9.2.10,φ(u)

uis piecewise continuous on 0, δ] and integrable. By Riemann’s Lemma

and Theorem 9.2.6, Sn(x) converges to S.

Example 9.2.1 Let f(x) be periodic with period 2π and

f(x) = x, −π < x < π.


By the formula of calculating the Fourier series, we get

f(x) ∼ 2 ·∞∑n=1

(−1)n+1 sinnx

n.

By Theorem 9.2.11, the Fourier series converges f(x) everywhere except at the points (2k + 1)π, k =

0,±1,±2, . . . . The Fourier series converges to 0 at these points.

Example 9.2.2 Let f(x) be periodic with period 2π and

f(x) = x2 − π < x < π.

By the formula of calculating the Fourier series we get

f(x) ∼ π2

3+ 4

∞∑n=1

(−1)n cosnx

n2.

By Theorem 9.2.11, the Fourier series converges to f(x) everywhere.

Note 2: sinnx and cosnx are odd function and even function respectively. If f(x) is odd function then

f(x) cosnx is still odd function, and

an =1

π

∫ π

−π

f(x) cosnx = 0.

Hence only the terms of sinnx appear in the Fourier series. Similarly if f(x) is even function then

f(x) sinnx is odd function, and all bn = 0. Hence only the terms of cosnx appear in the Fourier se-

ries. In above 2 examples, one is odd and one is even.

Example 9.2.3 The interval [−π, π] is divided into −π = x0 < x1 < · · · < xn = π, and f(x) is defined

as follows

f(x) = ci xi−1 < x < xi, i = 1, 2, . . . , n

and

f(x+ 2π) = f(x).

This kind of functions often appear in physics. By Theorem 9.2.11, the Fourier series of f(x) converges to

f(x) except at the points xi. At xi, i = 1, . . . , n− 1, the Fourier series converges to

f(xi + 0) + f(xi − 0)

2=ci+1 + ci

2.

At the points x0 and xn, the Fourier series converges to

f(xn + 0) + f(x0 − 0)

2=cn + c1

2.


9.3 Integration and Differentiation of Fourier Series

For a general function series, we usually require the uniform convergence of the series for termwise

integration and differentiation of the series. However, for Fourier series, the requirement of uniform con-

vergence is not necessary for termwise integration and differentiation of the series. This is actually the

advantage of the Fourier series. First, we introduce some lemmas.

Lemma 9.3.1 Suppose that f(x) is continuous and piecewise smooth on [a, b], then the First Fundamental

Theorem holds, i.e., ∫ b

a

f ′(x)dx = f(b)− f(a).

Proof. ∫ b

a

f ′(x)dx =

n∑k=1

∫ xk

xk−1

f ′(x)dx

=n∑

k=1

f(xk)− f(xk−1) = f(b)− f(a).

Lemma 9.3.2 Suppose that f(x) is continuous and piecewise smooth on [a, b], g(x) is continuous on [a, b],

andG(x) is an antiderivative of g(x). Then the Integration by Parts formula holds, i.e.,

∫ b

a

g(x)f(x)dx = G(x)f(x)|ba −∫ b

a

G(x)f ′(x)dx.

Proof. On [xk−1, xk], f(x) and G(x) are continuous, f ′(x) and g(x) are integrable. Then we have,

∫ b

a

g(x)f(x)dx =

n∑k=1

∫ xk

xk−1

f(x)g(x)dx

=

n∑k=1

[G(x)f(x)

∣∣∣xk

xk−1

−∫ xk

xk−1

g(x)f ′(x)dx]

= G(x)f(x)|ba −∫ b

a

G(x)f ′(x)dx.

Theorem 9.3.3 Suppose that f(x) is periodic with period 2π and piecewise continuous on [−π, π]. If

f(x) ∼ a02

+∞∑

n=1

(an cosnx+ bn sinnx),

then for any x ∈ (−∞,+∞), we have

∫ x

0

f(t)dt =a0x

2+

∞∑n=1

∫ x

0

(an cosnt+ bn sinnt)dt.


Proof. Let

F (x) =

∫ x

0

[f(t)− a0

2

]dt, x ∈ (−∞,+∞).

By Theorem 9.2.11, the Fourier series of F (x) converges to F (x), i.e.,

F (x) =A0

2+

∞∑n=1

(An cosnx+Bn sinnx), (9.3.1)

with

An =1

π

∫ π

−π

F (x) cosnxdx = −bnn, n = 1, 2, · · · ,

Bn =1

π

∫ π

−π

F (x) sinnxdx =ann, n = 1, 2, · · · .

Let x = 0 in (9.3.1), we get

A0 =

∞∑n=1

bnn.

Theorem 9.3.4 Suppose that f(x) is periodic with period 2π and continuous, and f ′(x) is piecewise smooth

on [−π, π]. If

f(x) ∼ a02

+∞∑

n=1

(an cosnx+ bn sinnx),

then for any x ∈ (−∞,+∞), we have

1

2[f ′(x+ 0) + f ′(x− 0)] =

∞∑n=1

(nbn cosnx− nan sinnx), (9.3.2)

i.e., the Fourier series of f ′(x) can be obtained from the Fourier series by termwise differentiation.

Proof. Let

f ′(x) ∼ a′02

+∞∑n=1

(a′n cosnx+ b′n sinnx).

Then we have

a′0 = 0, a′n = nbn, b′n − nan, n = 1, 2, · · · ,

and

f ′(x) ∼∞∑n=1

(nbn cosnx− nan sinnx).

Since f ′(x) is piecewise smooth, we have (9.3.2).

Next, we consider the uniform convergence of the Fourier series. The following lemma will be used in the

proof of the theorem.


Lemma 9.3.5 (Bessel’s Inequality) Suppose that f(x) is integrable on [a, b], ak and bk are the Fourier

coefficients of f(x). Then for any natural number n, we have

a202

+n∑

k=1

(a2k + b2k) ≤1

π

∫ π

−π

f2(x)dx.

Proof. Let

Sn(x) =a02

+n∑

k=1

(ak cos kx+ bk sinkx).

Then we have,

0 ≤ 1

π

∫ π

−π

[f(x)− Sn(x)]2dx

=1

π

∫ π

−π

f2(x)dx − a202

−n∑

k=1

(a2k + b2k).

Theorem 9.3.6 Suppose that f(x) is periodic with period 2π, continuous, and piecewise smooth on [−π, π].Then the Fourier series of f(x) converges uniformly to f(x) on (−∞,+∞).

Proof. By assumption,

f(x) =a02

+

∞∑k=1

(an cosnx+ bn sinnx). (9.3.3)

Let a′n and b′n be the Fourier coefficients of f ′(x), then we have,

f(x) =a02

+

∞∑k=1

(−b′n

ncosnx+

a′nn

sinnx).

From Bessel’s Inequality, the series∞∑k=1

(a2k + b2k) converges. Since

∣∣∣− b′nn

cosnx+a′nn

sinnx∣∣∣ ≤ 1

n2+ a2n + b2n,

(9.3.3) converges uniformly.

Theorem 9.3.7 (Parseval’s Equality) Suppose that the Fourier series of f(x) converges uniformly to f(x)

on [−π, π], then1

π

∫ π

−π

f2(x)dx =a202

+

∞∑n=1

(a2n + b2n).

Proof. Since [Sn(x) − f(x)]2 converges to 0 uniformly, then we have

limn→∞

1

π

∫ π

−π

[Sn(x)− f(x)]2dx = 0.


9.4 Fourier Expansion on Finite Interval

Suppose that we have a function f(x) defined on the interval [a, b], we want to obtain the Fourier series

of f(x). For a function f(x) defined in [a, b], it’s easy by a linear transform

x = ky + l

to obtain a function ϕ(y) = f(ky + l) such that ϕ(y) is defined on some standard interval such as

[0, 1], [−1, 1], [0, π], or [−π, π]. Hence, for simplicity, in the following discussion the definition domain

of function will be only [−π, π] and [0, π].

Fourier series of f(x) on [−π, π]: Let f(x) be defined on [−π, π]. We extend f(x) periodically to a

function defined on (−∞,+∞) through

f(x) = f(x+ 2π).

The original function f(x) may not satisfy the condition

f(−π) = f(π).

If the function values at these two end points are not important, then when making the periodical extension

we may revise both the function values at −π and π to be [f(−π+0)+ f(π− 0)]/2. This revision doesn’t

affect the calculation of Fourier coefficients. We calculate the coefficients of Fourier series by formula

an =1

π

∫ π

−π

f(x) cosnx, n = 0, 1, 2, . . .

bn =1

π

∫ π

−π

f(x) sinnx, n = 1, 2, . . .

and get

f(x) ∼ a02

+

∞∑n=1

(an cosnx+ bn sinnx).

However, we still need to check the convergence of the series. For example, if f(x) is piecewise smooth on

[−π, π]. Then we have

f(x+ 0) + f(x− 0)

2=

a02

+

∞∑n=1

(an cosnx+ bn sinnx), ∀x ∈ (−π, π),

f(−π + 0) + f(π − 0)

2=

a02

+

∞∑n=1

(an cosnx+ bn sinnx), x = −π, x = π.

Fourier series of f(x) on [0, π]: Let f(x) be defined on [0, π]. In this case, it is possible to make an odd or

even extension of f(x) to the interval [−π, 0] and then make a periodical extension to (−∞,+∞).


1. Even Extension: In this case, the extended function is even. As mentioned before, the Fourier series

only includes the terms of cosnx, i.e.,

f(x) ∼ a02

+

∞∑n=1

an cosnx.

The resulting series is called Fourier cosine series. Since f(x) cosnx is even,

an =1

π

∫ π

−π

f(x) cosnxdx

=2

π

∫ π

0

f(x) cosnxdx

For even extension, f(x) is continuous at the point 0,−π, and π.

2. Odd Extension: In this case, the extended function is Odd. As mentioned before, the Fourier series

only includes the terms of sinnx, i.e.,

f(x) ∼∞∑

n=1

bn sinnx.

The resulting series is called Fourier sine series. Since f(x) sinnx is even,

bn =1

π

∫ π

−π

f(x) sinnxdx

=2

π

∫ π

0

f(x) sinnxdx.

For odd extension, f(x) may not be continuous at the point 0, −π and π.

9.5 Exercise

1. Let f(x) = f(x+ 2π) and

f(x) =

⎧⎨⎩ 0, −π < x < 0,

1, 0 < x < π.

Find the Fourier series of f(x) and discuss its convergence.

2. Let g(x) = g(x+ 2π) and

g(x) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

0, −π < x < −π2 ,

1, −π2 < x < π

2 ,

0, π2 < x < π.

Find the Fourier series of g(x) and discuss its convergence.

3. Suppose that the Fourier series of f(x) converges to f(x) uniformly on [−π, π]. Prove the following

Parseval’s equality:1

π

∫ π

−π

f2(x)dx =a202

+∞∑n=1

(a2n + b2n).


4. Use Riemann Lemma to find the following limits:

(a) limλ→+∞

∫ π

−π

sin2 λxdx. (b) limλ→+∞

∫ a

0

cos2 λx

1 + xdx, a > 0.

5. Find the Fourier series and its sum for the following functions:

(a) f(x) =

⎧⎨⎩ 0, −π ≤ x < 0,

x, 0 ≤ x < π.

(b) f(x) =

⎧⎨⎩ x, −π ≤ x < 0,

x2, 0 ≤ x < π.

(c) f(x) = sinαx, −π ≤ x < π.

(d) f(x) = ex, −π ≤ x < π.

(e) f(x) =

⎧⎨⎩ 0, −π ≤ x < 0,

cosx, 0 ≤ x < π.

(f) f(x) =

⎧⎨⎩ 1, −π ≤ x < 0,

ex, 0 ≤ x < π.

(g) f(x) = max{2− |x|, 0}, −π ≤ x < π.

(h) f(x) = sgn(cosx), −π ≤ x < π.

6. Find the following sum by using a suitable Fourier series:

(a)∞∑

n=1

1

n2.

(b)∞∑

n=1

(−1)n+1

n2.

(c)∞∑

n=1

1

(2n− 1)2.

(d)∞∑

n=1

1

n4.

(e)∞∑

n=1

(−1)n+1

n4.

(f)∞∑

n=1

1

(2n− 1)4.

(g)∞∑

n=1

1

1 + n2.

(h)∞∑

n=1

(−1)n+1

1 + n2.

7. Suppose that f(x) is an odd function on (0, π). Also suppose that f(x) is non-negative and inte-

grable on (0, π). Prove that the Fourier coefficients of f(x) satisfies |bn| ≤ nb1, n = 1, 2, · · · .Hint. By induction, using Mean Value Theorem and sinnx = sin(n− 1 + 1)x.

8. Suppose that f(x) is integrable on [−π, π] and has the period π. Prove that the Fourier coefficients

of f(x) satisfies a2n−1 = 0, b2n−1 = 0, n = 1, 2, · · · .

9. Suppose that f(x) is integrable on [−π, π] and f(x+π) = −f(x). Prove that the Fourier coefficients

of f(x) satisfies a2n = 0, b2n = 0, n = 1, 2, · · · .

10. Find the sin and cos series of f(x) = π − x on [0, π].

11. Suppose that f(x) is continuous on [−π, π] and all the Fourier coefficients of f(x) are zero. Prove

that f(x) ≡ 0.

12. Suppose that f ′(x) is monotonically increasing and bounded on (0, 2π). Prove that the Fourier

coefficients of f(x) satisfies an ≥ 0, n = 1, 2, · · · .

Chapter 10

Improper Integral

10.1 Integration on an Unbounded Interval

In this section, we consider the first kind of improper integral and mainly the case with interval [a,+∞).

Definition 10.1.1 Suppose that f(x) is bounded on [a,+∞), and integrable on any interval [a, u], where

u > a. If

limu→+∞

∫ u

a

f(x)dx = A,

then we say that the improper integral of f(x) on [a,+∞) converges and denote by

∫ +∞

a

f(x)dx = A.

Otherwise, we say that the improper integral of f(x) on [a,+∞) diverges.

Example 10.1.1 Let ε > 0, then

∫ +∞

0

e−εxdx = limu→+∞

∫ u

0

e−εxdx = limu→+∞

[−e−εx

ε

]u0

=1

ε− lim

u→+∞ e−εu =1

ε.

Example 10.1.2 Let p > 1, then

∫ +∞

1

x−pdx = limu→+∞

∫ u

1

x−pdx = limu→+∞

[ x1−p

1− p

]u1

=1

p− 1− lim

u→+∞u1−p

p− 1=

1

p− 1.

195


10.1.1 Nonnegative Integrand

We first consider the case that the integrand f(x) is nonnegative.

Theorem 10.1.1 Suppose that f(x) ≥ 0 on [a,+∞) and integrable on any interval [a, u], where u > a.

Let

F (u) =

∫ u

a

f(x)dx.

Then∫ +∞

a

f(x)dx converges if and only if F (u) is bounded on [a,+∞).

Proof. Since f(x) ≥ 0,

F (u2)− F (u1) =

∫ u2

u1

f(x)dx ≥ 0, ∀u2 > u1.

Hence, F (u) is monotonically increasing on [a,+∞). By the Monotone Convergence Theorem,∫ +∞

a

f(x)dx = limu→+∞F (u)

exists if and only if F (u) is bounded on [a,+∞).

Theorem 10.1.2 (Comparison Test) Suppose that there is a number b, b > a, such that

0 ≤ f(x) ≤ Cg(x) ∀x ≥ b,

where C is a positive constant. We have

1. If∫ ∞

a

g(x)dx converges, then∫ ∞

a

f(x)dx converges.

2. If∫ ∞

a

f(x)dx diverges, then∫ ∞

a

g(x)dx diverges.

Proof.

1.∫ ∞

a

g(x)dx converges implies that∫ ∞

a

Cg(x)dx converges. From the assumption we get∫ u

b

f(x)dx ≤∫ ∞

b

Cg(x)dx. By Theorem 10.1.1,∫ ∞

b

f(x)dx converges and then∫ ∞

a

f(x)dx

converges.

2.∫ ∞

a

f(x)dx diverges implies that∫ u

a

f(x)dx is unbounded on a < u < +∞. By assumption,∫ u

a

g(x)dx is also unbounded on a < u < +∞. By Theorem 10.1.1,∫ ∞

a

g(x)dx diverges.

Theorem 10.1.3 Suppose that f(x) ≥ 0 and g(x) > 0 on [a,+∞) and

limx→+∞

f(x)

g(x)= l > 0.

Then∫ ∞

a

f(x)dx and∫ ∞

a

g(x)dx converge or diverge at the same time.

Chapter 10: Improper Integral 197

Proof. Let ε = l/2. Then from the assumption, there exists a M > 0 such that

l− ε <f(x)

g(x)< l + ε ∀x ≥M,

or,l

2g(x) < f(x) <

3l

2g(x) ∀x ≥M.

By Theorem 10.1.2,∫ ∞

a

f(x)dx and∫ ∞

a


Theorem 10.1.4 Suppose that f(x) ≥ 0 on [a,+∞), a > 0.

1. If for α > 1,

limx→+∞xαf(x) = l ≥ 0,

then∫ ∞

a

f(x)dx converges.

2. If for α ≤ 1,

limx→+∞xαf(x) = l > 0, or, lim

x→+∞xαf(x) = +∞,

then∫ ∞

a

f(x)dx diverges.

Proof.

1. From the assumption,

limx→+∞

f(x)

x−α= l,

and∫ ∞

a

x−αdx converges for α > 1. Then∫ ∞

a

f(x)dx converges from Theorem 10.1.3.

2. From the assumption,

limx→+∞

f(x)

x−α= l > 0,

and∫ ∞

a

x−αdx diverges for α ≤ 1. Then∫ ∞

a

f(x)dx diverges from Theorem 10.1.3. If

limx→+∞

f(x)

x−α= +∞, α ≤ 1,

then there exists a numberM such that

f(x) > x−α ∀x ≥M.

By Comparison Test,∫ ∞

a

f(x)dx diverges due to the divergence of∫ ∞

a

x−αdx when α ≤ 1.


10.1.2 Absolute Convergence Test

Consider a general improper integral∫ ∞

a

f(x)dx and let F (u) =

∫ u

a

f(x)dx. By Cauchy Conver-

gence Criterion, limu→+∞F (u) exists if and only if for given ε > 0, there exists a M > 0 such that

|F (u2)− F (u1)| < ε ∀u2 > u1 ≥M.

In other words, the following theorem holds.

Theorem 10.1.5∫ ∞

a

f(x)dx converges if and only if for given ε > 0, there exists a M > 0 such that

∣∣∣ ∫ u2

u1

f(x)dx∣∣∣ < ε ∀u2 > u1 ≥M.

Definition 10.1.6 If∫ ∞

a

|f(x)|dx converges, then we say that∫ ∞

a

f(x)dx converges absolutely. If∫ ∞

a

|f(x)|dx diverges, but∫ ∞

a

f(x)dx converges, then we say that∫ ∞

a

f(x)dx converges conditionally.

Theorem 10.1.7 If∫ ∞

a

f(x)dx converges absolutely, then∫ ∞

a

f(x)dx converges.

Theorem 10.1.8 (Absolute Convergence Test) Suppose that g(x) ≥ 0 on [a,+∞) and∫ ∞

a

g(x)dx con-

verges. If

|f(x)| ≤ Cg(x), x ≥M,

where C > 0 is a constant andM ≥ a, then∫ ∞

a

f(x)dx converges absolutely.

Theorem 10.1.9 (Abel Test) Suppose that∫ ∞

a

f(x)dx converges, g(x) is monotone and bounded on

[a,+∞). Then∫ ∞

a

f(x)g(x)dx converges.

Theorem 10.1.10 (Dirichlet Test) Suppose that∫ A

a

f(x)dx is bounded for any A ∈ [a,+∞), g(x) is

monotone on [a,+∞), and limx→+∞ g(x) = 0. Then

∫ ∞

a

f(x)g(x)dx converges.

10.1.3 Improper Integral on (−∞, b] and (−∞,+∞)

Definition 10.1.11 Suppose that f(x) is bounded on (−∞, b] and integrable on any interval [u, b], u < b.

If

limu→−infty

∫ b

u

f(x)dx = A.

Then we say that the improper integral of f(x) on (−∞, b] converges and denote it by∫ b

−∞f(x)dx = A.


A simple transform y = −x will change the integration interval from (−∞, b] to [−b,+∞), i.e.,

∫ b

−∞f(x)dx =

∫ +∞

−b

f(−y)dy.

Hence, all the theory above can be used for∫ b

−∞f(x)dx.

Definition 10.1.12 Suppose that f(x) is bounded on (−∞,+∞) and integrable on any interval [u, v]. If

limu→−∞,v→+∞

∫ v

u

f(x)dx = A,

or equivalently,

limu→−∞

∫ 0

u

f(x)dx+ limv→+∞

∫ v

0

f(x)dx = A.

Then we say that the improper integral of f(x) on (−∞,+∞) converges and denote it by

∫ +∞

−∞f(x)dx = A.

By the definition,∫ +∞

−∞f(x)dx converges if and only if both

∫ 0

−∞f(x)dx and

∫ +∞

0

f(x)dx converge

and ∫ +∞

−∞f(x)dx =

∫ 0

−∞f(x)dx +

∫ +∞

0

f(x)dx.

10.2 Integral with an Unbounded Integrand

In this section, we discuss another kind of improper integral: the interval is finite, but the function is

unbounded.

Definition 10.2.1 Suppose that f(x) is unbounded on [a, b] and limx→b

f(x) = +∞ (or −∞), but f(x) is

bounded and integrable on any interval [a, u], a < u < b, and

limu→b

∫ u

a

f(x)dx = A.

Then we say that the improper integral of f(x) on [a, b] converges and denote it by

∫ b

a

f(x)dx = A.

If the limit does not exist, we say that the improper integral of f(x) on [a, b] diverges.


10.2.1 Nonnegative Integrand

Parallel to Theorem 10.1.1 to Theorem 10.1.4, we have the following theorems.

Theorem 10.2.1 Suppose that f(x) ≥ 0 on [a, b] and integrable on any interval [a, u], where a < u < b.

Let

F (u) =

∫ u

a

f(x)dx.

Then∫ b

a

f(x)dx converges if and only if F (u) is bounded on [a, b).

Theorem 10.2.2 (Comparison Test) Suppose that

0 ≤ f(x) ≤ Cg(x) ∀a ≤ x < b,

where C is a positive constant. We have

1. If∫ b

a

g(x)dx converges, then∫ b

a

f(x)dx converges.

2. If∫ b

a

f(x)dx diverges, then∫ b

a

g(x)dx diverges.

Theorem 10.2.3 Suppose that f(x) ≥ 0 and g(x) > 0 on [a, b), and

limx→b

f(x) = +∞, limx→b

g(x) = +∞.

If

limx→b

f(x)

g(x)= l > 0,

then∫ b

a

f(x)dx and∫ b

a


Theorem 10.2.4 Suppose that f(x) ≥ 0 on [a, b).

1. If for α < 1,

limx→b

(b − x)αf(x) = l ≥ 0,

then∫ b

a

f(x)dx converges.

2. If for α ≥ 1,

limx→b

(b− x)αf(x) = l > 0, or, limx→b

(b− x)αf(x) = +∞,

then∫ b

a

f(x)dx diverges.


10.2.2 Absolute Convergence Test

For general improper integrals on [a, b], we have the following theorems which are parallel to Theorem

10.1.5, Theorem 10.1.7, and Theorem 10.1.8.

Theorem 10.2.5 Suppose that limx→b

f(x) = +∞ (or − ∞). Then∫ b

a

f(x)dx converges if and only if for

given ε > 0, there exists a δ > 0 such that∣∣∣ ∫ u2

u1

f(x)dx∣∣∣ < ε ∀b− δ < u1 < u2 < b.

Definition 10.2.6 If∫ b

a

|f(x)|dx converges, then we say that∫ b

a

f(x)dx converges absolutely. If∫ b

a

|f(x)|dx diverges, but∫ b

a

f(x)dx converges, then we say that∫ b

a

f(x)dx converges conditionally.

Theorem 10.2.7 If∫ b

a

f(x)dx converges absolutely, then∫ b

a

f(x)dx converges.

Theorem 10.2.8 (Absolute Convergence Test) Suppose that g(x) ≥ 0 on [a, b) and∫ b

a

g(x)dx converges.

If

|f(x)| ≤ Cg(x), a′ ≤ x < b,

where C > 0 is a constant and a ≤ a′ < b, then∫ b

a

f(x)dx converges absolutely.

10.2.3 Other Cases

Definition 10.2.9 Suppose that f(x) is unbounded on [a, b] and limx→a

f(x) = +∞ (or −∞), but f(x) is

bounded and integrable on any interval [u, b], a < u < b. If

limu→a

∫ b

u

f(x)dx = A,

then we say that the improper integral of f(x) on [a, b] converges and denote it by∫ b

a

f(x)dx = A.

If the limit does not exist, we say that the improper integral of f(x) on [a, b] diverges.

Definition 10.2.10 Suppose that f(x) is unbounded on [a, b] and

limx→a

f(x) = ±∞, limx→b

f(x) = ±∞,

but f(x) is bounded and integrable on any interval [u, v], a < u < v < b. If

limu→a,v→b

∫ v

u

f(x)dx = A,


or equivalently,

limu→a

∫ c

u

f(x)dx+ limv→b

∫ v

c

f(x)dx = A, a < c < b,

then we say that the improper integral of f(x) on [a, b] converges and denote it by∫ b

a

f(x)dx = A.

By the definition,∫ b

a

f(x)dx converges if and only if both∫ c

a

f(x)dx and∫ b

c

f(x)dx converge (a < c <

b) and ∫ b

a

f(x)dx =

∫ c

a

f(x)dx+

∫ b

c

f(x)dx.

10.3 Gamma Function and Beta Function

In this section, we recommend two important functions that are defined by improper integrals. They

are widely used in physics, statistics, as well as other branches of mathematics.

10.3.1 Gamma Function

The following improper integral that includes a parameter s

Γ(s) =

∫ ∞

0

xs−1e−xdx, s > 0

is called Gamma Function. When s < 1, limx→0

xs−1e−x = +∞. Hence it is of two kinds of improper

integrals. However, both integrals∫ 1

0

xs−1e−xdx and∫ ∞

1

xs−1e−xdx converge for s > 0. Therefore,

Gamma Function is well defined on the interval (0,+∞).

Theorem 10.3.1 For s > 0,

Γ(s+ 1) = sΓ(s).

Theorem 10.3.2 For natural number n,

Γ(n+ 1) = n!.

10.3.2 Beta Function

Beta function β(p, q) involves two parameters, and is defined by the improper integral

β(p, q) =

∫ 1

0

xp−1(1 − x)q−1dx, p > 0, q > 0.


When 0 < p < 1 and 0 < q < 1, the integrand is unbounded on (0, 1) and

limx→1

xp−1(1− x)q−1 = +∞, limx→0

xp−1(1 − x)q−1 = +∞.

We can write the improper integral in form of

β(p, q) =

∫ 1/2

0

xp−1(1− x)q−1dx +

∫ 1

1/2

xp−1(1− x)q−1dx.

Since

xp−1(1− x)q−1 ≤(12

)q−1

xp−1, 0 < x ≤ 1

2,

xp−1(1− x)q−1 ≤(12

)p−1

xq−1,1

2≤ x < 1.

By Comparison Test, when 0 < p < 1 and 0 < q < 1, both integrals converge. Hence the Beta function is

defined for p > 0, q > 0.

Theorem 10.3.3 β(p, q) = β(q, p).

Theorem 10.3.4 β(p+ 1, q + 1) =q

p+ q + 1β(q + 1, p) =

p

p+ q + 1β(p, q + 1).

Theorem 10.3.5 β(p, q) =Γ(p)Γ(q)

Γ(p+ q), p > 0, q > 0.

10.4 Exercise

1. Evaluate the following integrals:

(a)∫ +∞

0

dx

1 + x3. (b)

∫ +∞

0

e−ax sinxdx, a > 0.

2. Evaluate the following integrals:

(a)∫ 1

0

dx

(2 − x)√1− x

. (b)∫ 1

0

sin−1 x

xdx.

3. Discuss the convergence of the following integrals:

(a)∫ +∞

1

dx√1 + x2 ln2(1 + x)

. (b)∫ +∞

0

xp

1 + xqdx, p > 0, q > 0.

4. Discuss the convergence of the following integrals:

(a)∫ 1

0

lnx

1− xdx. (b)

∫ 1

0

xp

lnq(1 + x)dx, p > 0, q > 0.


ë�©z[1] P.M. Fitzpatrick, Advanced Calculus, Thomson Higher Education, 2006.

[2] Wilfred Kaplan, Advanced Calculus, Addison-Wesley, 1993.

[3] J.R. Kirkwood, An Introduction to Analysis, 2nd edition, PWS-KENT, 1995.

[4] Jonathan Lewin, An Introduction to Mathematical Analysis, 2nd edition, McGraw

Hill, 1993.

[5] Paulo Ney De Souza and Jorge-Nuno Silva, Berkeley Problems in Mathematics,

3rd edition, Springer, 2004.

[6] G. Klambauer, Mathematical Analysis, Marcel Dekker Inc, 1975.

[7] Walter Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw Hill,

1976.

[8] �â7x�℄, �È©Æ�§, p��Ñ��, 18�, 2006.

[9] 3�õ�Û, êÆ©ÛSK8µJ«§)Kg´§�Y, ¤½z?�, H��"�,ìÀ�ÆEâÑ��, 2007.

[10] o¤Ù, ��¬, êÆ©Û, �ÆÑ��, 2007.

[11] �¨¬, �g�, ´{|, a½>, êÆ©ÛµSK�ùÂ, p��Ñ��, 2003.

[12] �7, êÆ©Û²;SK�)Û,p��Ñ��, 2004.

205

Documents

Mathematical Analysis - Hong Kong Baptist Universityxwu/m1111/MathAnal.pdf8 Mathematical Analysis. Chapter 1 Sets and Numbers 1.1 Sets Set theory is the foundation in which virtually