Introduction to Mathematical Analysis

Lecture Note

Introduction to Mathematical Analysis

0 0.2 0.4 0.6 0.8 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

FIRST SEMESTER 2010

Department of Mathematics

The College of Natural Sciences

Kookmin University

COPYRIGHT ©2010 DEPARTMENT OF MATHEMATICS, KOOKMIN UNIVERSITY. ALL RIGHTS RESERVED.

Lecture note forIntroduction to Mathematical AnalysisDepartment of MathematicsThe College of Natural SciencesKookmin University861-1, Jeongneung-dong, Seongbuk-guSeoul, 136-702, Koreahttp://math.kookmin.ac.kr

TABLE DES MATIÈRES 3

Table des matières

1 The Real Number System 4

1.1 Principle of Mathematical Induction . . . . . . . . . . . . . . . . . . . . 4

1.2 The Algebraic Properties of Real Number R . . . . . . . . . . . . . . . . 5

1.3 The Order Properties of Real Number R . . . . . . . . . . . . . . . . . . 6

1.4 The Completeness Property of Real Number R . . . . . . . . . . . . . . . 10

1.5 Exercises for Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Sequences 14

2.1 Convergent Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2 Limit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3 Monotone Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.4 Subsequences and the Cauchy criterion . . . . . . . . . . . . . . . . . . . 27

2.5 Upper and Lower Limits of Bounded and Unbounded Sequences . . . . . 35


3 Limits of Functions 42

3.1 Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.2 Some Properties of Limits of Functions . . . . . . . . . . . . . . . . . . . 50


4 Continuous Functions 62

4.1 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

4.2 Properties of Continuous Functions . . . . . . . . . . . . . . . . . . . . . 68

4.3 Uniformly Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . 73


References 85

4 The Real Number System


The pain purpose of this chapter is presentation of basic background for the studyof mathematical analysis.

1.1 Principle of Mathematical Induction

Mathematical Induction is one of powerful method of proof that is frequently usedto establish the validity of statements that are given in terms of the natural numbers.Although its utility is restricted to this rather special context, mathematical inductionis an indispensable tool in all branches of mathematics. In this section, we state theprinciple and give various examples to illustrate how inductive proofs proceed.

Let us denote N be the set of natural numbers :

N = {1, 2, 3, · · · } ,

with the usual operations of addition and multiplication, and with the meaning of a na-tural number being less than another one. We will also assume the following fundamentalproperty of natural number.

Axiom 1.1 (Well-ordering property) Every nonempty subset of N has a least ele-ment.

A more detailed statement of this property is as follows : If S is a subset of N and ifS 6= φ, then there exists m ∈ S such that m ≤ k for all k ∈ S. Based on this property,the principle of mathematical induction can be expressed in terms of subsets of N.

Theorem 1.2 (Principle of Mathematical Induction) Let S be a subset of N thatsatisfies the following two properties :

1. The number 1 ∈ S.

2. For every k ∈ N, if k ∈ S then k + 1 ∈ S.

Then S = N.

Now, let us generalize the principle of mathematical induction. Let us denote P (n)be a meaningful statement about n ∈ N. Then P (n) may be true for some values n andfalse for others. With this statement, the principle of mathematical induction can bestated as follows :

Theorem 1.3 For each n ∈ N, let P (n) be a statement about n. Suppose that

1. P (1) is true.

2. For every k ∈ N, if P (k) is true then P (k + 1) is true.

Then P (n) is true for all n ∈ N.

1.2 - The Algebraic Properties of Real Number R 5

Example 1.4 Use induction to show that

12 + 22 + 32 + · · ·+ n2 =1

6n(n + 1)(2n + 1)

for every n ∈ N.

In fact, it may happen that statement P (n) are false for some n ∈ N but then are truefor every n ≥ n0 for some particular n0. Then the principle of mathematical inductioncan be modified to deal with this situation as follows :

Theorem 1.5 (Principle of Mathematical Induction (second version)) Letn0 ∈ N and P (n) be a statement for each natural number n ≥ n0. Suppose that

1. P (n0) is true.2. For every k(∈ N) ≥ n0, P (k) is true implies P (k + 1) is true.

Then P (n) is true for all n ≥ n0.

It is worth mentioning that another version of the principle of mathematical induc-tion so called Principle of Strong Induction is sometimes quite useful. It can be statedas follows :

Theorem 1.6 (Principle of Strong Induction) Let S be a subset of N that satisfiesthe following two properties :

1. 1 ∈ S.2. For every k ∈ N, if {1, 2, · · · , k} ⊆ S then k + 1 ∈ S.

Then S = N.

1.2 The Algebraic Properties of Real Number R

In this section, we shall give the algebraic structure of the real number system. Brieflyexpressed, the real numbers form a field in the sense of abstract algebra. We shall nowexplain what that means. We begin with a definition of binary operation.

Definition 1.7 (Binary operation) A binary operation (or simply, operation) Bon a set F is a function from F × F into F .

In the set R of real numbers, there are two binary operations (denoted by + and· and called addition and multiplication, respectively) satisfying the following familiarproperties :(A1) a + b = b + a for all a, b ∈ R (commutative property of addition)(A2) (a + b) + c = a + (b + c) for all a, b, c ∈ R (associative property of addition)(A3) There exists an element 0 ∈ R such that a+0 = 0+ a = a for all a ∈ R (existence

of a zero element)(A4) For each a ∈ R, there exists an element −a ∈ R such that a + (−a) = 0 and

(−a) + a = 0 (existence of negative element)


(M1) a · b = b · a for all a, b ∈ R (commutative property of multiplication)(M2) (a · b) · c = a · (b · c) for all a, b, c ∈ R (associative property of multiplication)(M3) There exists an element 1 ∈ R such that a · 1 = 1 · a = a for all a ∈ R (existence

of a unit element)(M4) For each nonzero a ∈ R, there exists an element 1

a∈ R such that a ·

(1a

)= 1 and(

1a

)· a = 1 (existence of reciprocals)

(D) a · (b + c) = (a · b) + (a · c) and (b + c) · a = (b · a) + (c · a) for all a, b, c ∈ R(distributive property of multiplication over addition).

From now on, we will obtain some corresponding results of them. First, we will showthat 0 and 1 is the only element of R that satisfies (A3) and (M3), respectively.

Theorem 1.8 Let a, u and z are elements of R1. If a + z = a then z = 0.2. For a 6= 0, if a · u = a then u = 1.

Second, we will show that −a and 1a

(when a 6= 0) are uniquely determined by theproperties given in (A4) and (M4), respectively.

Theorem 1.9 Let a and b are elements of R,1. If a + b = 0 then b = −a.2. For a 6= 0, if a · b = 1 then b = 1

a.

Third, now we can obtain the following uniqueness of solution of equations :

Theorem 1.10 Let a and b are elements of R,1. The equation a + x = b has the unique solution x = (−a) + b.2. For a 6= 0, the equation a · x = b has the unique solution x =

(1a

)· b.

Last, we would like introduce some properties :

Theorem 1.11 Let a and b are elements of R then1. a · 0 = 0.2. (−a) · (−b) = ab.

Note that one can explore more algebraic properties of real number. We recommendsome references [2, 3, 4, 5, 7].

1.3 The Order Properties of Real Number R

In this section, we introduce the important order properties of real number R, whichwill play a very important role in subsequent sections. The simplest way to introducethe notion of order is to make use of the notion of strict positivity, which we now explain

1.3 - The Order Properties of Real Number R 7

Axiom 1.12 (Axiom of order) A relation < defined on R×R satisfies the followingaxiom of order

1. For a, b ∈ R, exactly one of the following holds (property of trichotomy) :

a = b, a < b or b < a.

2. For a, b ∈ R, if 0 < a and 0 < b then 0 < a + b and 0 < ab.3. For a, b, c ∈ R, if a < b then a + c < b + c.

If a ∈ R, we say that a is a strictly positive real number and write a > 0. Ifa is either in R or is 0, we say that a is a positive real number and write a ≥ 0. If−a ∈ R, we say that a is a strictly negative real number and write a < 0. If −a iseither in R or is 0, we say that a is a negative real number and write a ≤ 0.

Now, we introduce some well-known properties

Theorem 1.13 Let a, b, c ∈ R then1. If a < b then −b < −a.2. If a < b and b < c then a < c.3. a2 ≥ 0 therefore 1 > 0.4. If a < b and c < 0 then bc < ac.5. If 0 < a then 0 < 1

a.

6. If 0 < a < b then 0 < 1b

< 1a.

Based on these properties, one can prove following :

Theorem 1.14 For a, b ∈ R, if a < b then

a <1

2(a + b) < b.

Corollary 1.15 For 0 < a ∈ R,

0 <1

2a < a.

The corollary 1.15 implies that for any strictly given positive number a, there is anotherstrictly smaller and strictly positive number (for example, 1

2a). Thus there is no smallest

strictly positive real number greater than 0. This observation leads to the next result,which will be used frequently as a method of proof.

Corollary 1.16 If a ∈ R satisfies 0 ≤ a < ε for every ε > 0 then a = 0.

It is worth mentioning that in order to prove that a number a ≥ 0 is actually equal tozero, it suffices to show that a is smaller than an arbitrary positive number.

From the trichotomy property in axiom 1.12 assures that if a 6= 0, then one of thenumbers a and −a is strictly positive. The absolute value of a 6= 0 is defined to be thestrictly positive one of the pair {a,−a}, and the absolute value of 0 is defined to be 0.


Definition 1.17 If a ∈ R, the absolute value of a is denoted by |a| and is defined by

|a| ={

a if a ≥ 0−a if a < 0.

Based on the definition 1.17, we can observe that |a| ≥ 0 for all a ∈ R. It meansthat |a| = 0 if and only if a = 0. Moreover, | − a| = |a| for all a ∈ R. Some additionalproperties are as follows :

Theorem 1.18 Let a, b ∈ R then

1. |ab| = |a||b|.2. For 0 < c ∈ R, |a| ≤ c if and only if −c ≤ a ≤ c.

3. −|a| ≤ a ≤ |a|.

Now, we will consider the following important and famous inequality will be usedfrequently :

Theorem 1.19 (Triangle inequality) If a, b ∈ R then

|a + b| ≤ |a|+ |b|.

It can be shown that equality occurs in the triangle inequality if and only if ab > 0,which is equivalent to saying that a and b have the same sign. There are many variationsof the triangle inequality. Herein, we consider two of them.

Corollary 1.20 If a, b ∈ R then

1. ||a| − |b|| ≤ |a− b|.2. |a− b| ≤ |a|+ |b|.

The following corollary is the generalized triangle inequality :

Corollary 1.21 If a1, a2, · · · , an ∈ R then

|a1 + a2 + · · ·+ an| ≤ |a1|+ |a2|+ · · ·+ |an|.

Now, let us mention a simple but important thing. We will later need precise languageto discuss the notion of one real number being close to another. If a is given real number,then saying that a real number b is close to a should mean that the distance |a − b|between them is small. A context in which this area can be discussed is provided by theterminology of neighborhoods, which we now define.

Definition 1.22 Let a ∈ R and ε > 0. The ε−neighborhood of a is the set

Nε(a) := {x ∈ R : |x− a| < ε} .

1.3 - The Order Properties of Real Number R 9

With this definition and corollary 1.16, we can obtain the following important theo-rem.

Theorem 1.23 Let a, b ∈ R. For arbitrary ε > 0, if |a− b| < ε then a = b.

The order relation on real number R determines a natural collection of subsets calledintervals. The following notations and terminology for these special sets will be familiarfrom earlier courses.

Definition 1.24 Let a, b ∈ R satisfy a < b

1. The open interval determined by a and b is the set

(a, b) := {x ∈ R : a < x < b} .

2. The closed interval determined by a and b is the set

[a, b] := {x ∈ R : a ≤ x ≤ b} .

3. The half-open (or half-closed) intervals determined by a and b is the set

[a, b) := {x ∈ R : a ≤ x < b}(a, b] := {x ∈ R : a < x ≤ b} .

Notice that the points a and b are called the endpoints of the interval.

There are five types of unbounded intervals for which the symbols ∞(or +∞) and−∞1 are used as notational convenience in place of the endpoints. The infinite openintervals are the sets of the form

(a,∞) := {x ∈ R : a < x}(−∞, b) := {x ∈ R : x < b} .

Notice that the first and second sets have no upper and lower bounds, respectively.Adjoining endpoint gives the infinite closed intervals as

[a,∞) := {x ∈ R : a ≤ x}(−∞, b] := {x ∈ R : x ≤ b} .

It is often convenient to think of the entire set R as an infinite interval. In this case, wewrite

(−∞,∞) := R.

An obvious property of intervals is that if two points a, b with a < b belong to aninterval I then any point lying between them also belongs to I. In other words, if a andb belongs to I then the interval [a, b] is contained in I.

Theorem 1.25 (Characterization theorem) If I is a subset of R that contains atleast two points a and b and a < b. If every t satisfies a < t < b belongs to I then I isan interval.

1It must be emphasized that ∞ and −∞ are not elements of R, but only convenient symbols.


1.4 The Completeness Property of Real Number R

In this section we shall present an important property of the real number systemwhich is often called the completeness property since it guarantees the existence ofelements in R when certain hypotheses are satisfied.

We now introduce the notion of an upper bound of a set of real numbers.

Definition 1.26 Let X be a nonempty subset of R.1. The set X is said to be bounded above if there exists a number a ∈ R such that

x ≤ a for all x ∈ X. Each number a is called an upper bound of X.2. The set X is said to be bounded below if there exists a number b ∈ R such that

b ≤ x for all x ∈ X. Each number b is called an lower bound of X.3. The set X is said to be bounded if it is both bounded above and bounded below.

Example 1.27 The set

A =

{1− 1

n: n = 1, 2, 3, · · ·

}is bounded below. The number 0 and any number smaller than 0 is a lower bound of A.This set is also bounded above. The number 1 and any number larger than 1 is an upperbound.

If a set has one upper bound then it has infinitely many upper bounds, because ifa is an upper bound of X then the numbers a + 1, a + 2, · · · are also upper boundsof X (similarly, lower bound is also). So, in the set of upper bounds of X and set oflower bounds of X, we focus on their least and greatest elements, respectively, for specialattention in the following definition.

Definition 1.28 Let X be a nonempty subset of R1. If X is bounded above then a number a is said to be a supremum or a least

upper bound of X if it satisfies the following conditions :(a) a is an upper bound of X

(b) if b is any upper bound of X then a ≤ b.2. If X is bounded above then a number b is said to be a infimum or a greatest

lower bound of X if it satisfies the following conditions :(a) b is an lower bound of X

(b) if a is any lower bound of X then a ≥ b.

If the supremum or the infimum of a set X exists, we will denote them by supX andinfX. Let us note the for a nonempty subset X of R,

1. a ∈ X is the maximum of X if x ≤ a for every x ∈ X and denote

a = max X.

1.4 - The Completeness Property of Real Number R 11

2. b ∈ X is the minimum of X if b ≤ x for every x ∈ X and denote

b = min X.

If X contains maximum then maxX=supX. Similarly, if X contains minimum thenminX=infX.

Theorem 1.29 Let A be a bounded above, nonempty subset of R and a ∈ R is an upperbound of A. Then the following statements are equivalent :

1. a is the supremum of A

2. for any b ∈ R satisfying b < a, there exists x ∈ A such that b < x ≤ a.

It is impossible to prove on the basis of the field and order properties of real numberthat every nonempty subset of R that is bounded above has a supremum in R. However,it is a deep and fundamental property of the real number system that this is indeedthis case. We will make frequent and essential use of this property, especially in ourdiscussion of limiting processes. The following statement concerning the existence ofsuprema is our final assumption about R.

Axiom 1.30 (Completeness property of real number) Every nonempty set ofreal numbers which has an upper bound also has a supremum in R.

This property is also called the supremum property of real number. The analogousproperty for infima can be deduced from the completeness property as follows :

Theorem 1.31 Every nonempty set of real numbers which has a lower bound has aninfimum in R.

So, based on the completeness property of R, we can say that R is a complete or-dered (field). From now on, we will give some important applications in order to derivefundamental properties of R.

One important consequence of the supremum property is that the set of naturalnumbers N is not bounded above in R.

Theorem 1.32 (Archimedean property) If x ∈ R, there is a natural number nx ∈N such that

x < nx.

This property induces following corollary.

Corollary 1.33 Let x, y be real numbers.

1. If x > 0 then there exist n ∈ N such that

y < nx.


2. For any x > 0, there exist n ∈ N such that

0 <1

n< x.

3. For any x > 0, there exist n ∈ N uniquely such that

n ≤ x < n + 1.

One important property of the supremum property if that it assures the existence ofcertain real numbers. We shall make use of it many times in this way. At the moment wewill show that is guarantees the existence of a positive real number x such that x2 = 2,that is, a positive square root of 2.

Theorem 1.34 There exists a positive number x ∈ R such that x2 = 2.

From the above theorem, we now know that there exists at least one irrational realnumber, namely

√2. Actually, there are more irrational numbers than rational numbers

in the sense that the set of rational numbers is countable, while the set of irrationalnumbers is uncountable (as shown in the Set Theory). However, we will show that inspite of this apparent disparity, the set of rational numbers is dense in R in the sensethat given any two real numbers there is a rational number between them (in fact, thereare infinitely many such rational numbers).

Theorem 1.35 (The density theorem) If x and y are any real numbers with x < y.1. Then there exists a rational number r such that x < r < y.2. Then there exists a irrational number z such that x < z < y.

Another method of completing the rational numbers to obtain R was revised byDedekind. It is based on the notion of a cut.

Definition 1.36 An ordered pair (A, B) of non-empty subset of R is said to form a cutif

A ∩B = φ, A ∪B = R and a < b

for all a ∈ A and b ∈ B.

Example 1.37 A typical example of a cut in R is obtained for a fixed element α ∈ Rby defining

A = {x ∈ R ≤ α} and B = {x ∈ R > α} .

Alternatively, we could take

A = {x ∈ R < α} and B = {x ∈ R ≥ α} .

Actually, what Dedekind did was, in essence, to define a real number to be a cut inthe rational number system. This procedure enables one to construct the real numbersystem R from the set of rational numbers.

Theorem 1.38 (Dedekind cur theorem) If (A, B) is a cut in R then there exists aunique number α ∈ R such that a ≤ α for all a ∈ A and α ≤ b for all b ∈ B.

1.5 - Exercises for Chapter 1 13

1.5 Exercises for Chapter 1

1. Prove that n! > 2n for all n ≥ 4, n ∈ N.2. If a ∈ R and a 6= 0, prove that

−1

a=

1

−a.

a

a= 1.

3. If a, b, c, d ∈ R, prove that(a) if b 6= 0 and d 6= 0 then (a

b

) ( c

d

)=

ac

bd.

(b) if b 6= 0 and d 6= 0 thena

b+

c

d=

ad + bc

bd.

4. If a1, a2, · · · , an ∈ R then

|a1 + a2 + · · ·+ an| ≤ |a1|+ |a2|+ · · ·+ |an|.

5. Prove the Bernoulli’s inequality : If x > −1 then

(1 + x)n ≥ 1 + nx.

6. Obtain the supremum and infimum of following sets :

S1 =

{1

n− (−1)n : n ∈ N

}.

S2 =

{1 +

(−1)n

n: n ∈ N

}.

S3 = {x ∈ R : |2x− 1| < 11} .

S4 =

{(−1)nn

2n + 1: n ∈ N

}.

7. Prove corollary 1.16 by using the completeness property of real number.

14 Sequences

2 Sequences

This chapter will deal primarily with sequences of real numbers. We shall begin witha study of the convergence of sequences. Some of the results in this chapter may befamiliar to the students from other courses, e.g. Calculus, but the study here is intendedto be rigorous and to give certain more profound results than are usually discussed inearlier courses.

2.1 Convergent Sequences

We begin our study with the introduction of a sequence of real numbers.

Definition 2.1 (A sequence of real numbers) A sequence of real numbers (ora sequence in R) is a function defined on the set N = {1, 2, · · · } of natural numberswhose range is contained in the set R of real numbers.

In other words, a sequence in R assigns to each natural number n = 1, 2, · · · auniquely determined real number. If f : N −→ R is a sequence, we will usually denotethe value of f at n by the symbol f(n) := xn. The values xn are called the (n−th) termsor the elements of the sequence. We will denote this sequence by the notations {xn}∞n=1

or simply {xn}.

Example 2.2 Let us consider the sequence

xn = (−1)n.

This sequence has infinitely many terms that alternate between −1 and 1, whereas theset of values {xn} is equal to the set {−1, 1}.

Let us consider the sequence whose n−th terms is defined by the formula

xn = 1 +1

2n.

The first four terms of this sequence are3

2,5

4,9

8,17

16

and the terms corresponding to n = 40, 41, 42 are1099511627777

1099511627776,2199023255553

2199023255552,4398046511105

4398046511104

which are close to 1. For example, x40 = 10995116277771099511627776

differs from 1 by only 11099511627776

≈9.1× 10−13. It is clear that xn is close to 1 for all large enough positive integers n. Forthis reason we can say that the sequence xn has limit 1, refer to Fig. 2.1.

Generally, we say that a sequence {xn} has limit L if xn is close to L for all largepositive integers n. To define the limit of a sequence, we need to make the concepts closeto and for all large positive integers n precise. In fact, there are a number of differentlimit concepts in real analysis. In this chapter, we introduce the following definition oflimit by using theorem 1.23 in chapter 1.

2.1 - Convergent Sequences 15

0 5 10 15 20 25 30 35 400.9

1

1.1

1.2

1.3

1.4

1.5

1.6

Fig. 2.1 – First 10 values of sequence xn = 1 + 12n . xn is getting close to 1 when n is

increasing.

Definition 2.3 (Convergent and limit) A sequence {xn} in R is said to convergeto L ∈ R or L is said to be a limit of {xn}, if for every ε > 0 there exists a naturalnumber N(ε) such that for all n ≥ N(ε), the terms xn satisfy

|xn − L| < ε.

If a sequence has a limit, we say that the sequence if convergent ; if it has no limit, wesay that the sequence is divergent.

1. Let us notice that the notation N(ε) is used to emphasize that the choice of Ndepends on the value of ε. However, it is often convenient to write N instead ofN(ε). For the sake of simplicity, we will use N instead of N(ε).

2. When a sequence {xn} has limit L, we will use the notation

limn→∞

xn = L, limn

xn = L or lim xn = L.

3. Sometimes, the symbolism xn −→ L is used in order to indicate the intuitive ideathat the values xn approach the number L as n −→∞.

Example 2.4 A sequence

{xn} =

{1

n: n ∈ N

}is converges to 0. Because, if ε > 0 is given then 1

ε> 0. By the archimedean property (see

theorem 1.32 in chapter 1), there exists a natural number N = N(ε) such that 1N

< ε.Then, if n ≥ N , we have 1

n≤ 1

N< ε. Consequently, if n ≥ N then∣∣∣∣ 1n − 0

∣∣∣∣ =1

n≤ 1

N< ε.

Therefore, we can say that the sequence {xn} converges to 0.

16 Sequences

Example 2.5 A sequence

{xn} =

{2 +

1

2n: n ∈ N

}is converges to 2.

Proof. Let ε > 0 be given. In order to find N , we first note that if n ∈ N and a > −1then by applying Bernoulli’s inequality,

1

(1 + a)n≤ 1

1 + na<

1

n.

Now choose N such that 1N

< ε. Then n ≥ N implies that∣∣∣∣(2 +1

2n

)− 2

∣∣∣∣ =1

2n≤ 1

1 + n<

1

n≤ 1

N< ε.

Hence, we have shown that the limit of the sequence is 2.

The next theorem allows us to speak of the limit of a sequence. This is a simple butimportant property of limit of sequence.

Theorem 2.6 (Uniqueness of limits) The limit of a sequence in R is unique. Thatis, if a sequence {xn} has limit L1 and L2 then L1 = L2.

Proof. Suppose that L1 and L2 are both limits of {xn}. Then for each ε > 0 there existsN1 such that for all n ≥ N1,

|xn − L1| <ε

2.

Moreover, there exists N2 such that for all n ≥ N2,

|xn − L2| <ε

2.

We let N be the larger of N1 and N2, i.e., N = max {N1, N2}, then for n ≥ N we applythe triangle inequality (theorem 1.19 in chapter 1) to obtain

|L1 − L2| = |L1 − xn + xn − L2|

≤ |L1 − xn|+ |xn − L2| <ε

2+

ε

2= ε.

Since ε > 0 is an arbitrary positive number, we conclude that L1 = L2 by theorem 1.23.

Let us notice that, above theorem can be argued by contradiction. A more detaineddescription, see [4, Theorem 10.3].

Now, we will consider some results that enable us to evaluate the limits of certain se-quences of real numbers. These results will expand our collection of convergent sequencesrather extensively. We begin by establishing an important property of convergent se-quences that will be needed in this and later sections.

Definition 2.7 (Bounded sequences) Let {xn} be a sequence of real numbers.

2.2 - Limit Theorems 17

1. {xn} is said to be bounded above if there exists a real number M > 0 such thatfor all n ∈ N,

xn ≤ M.

2. {xn} is said to be bounded below if there exists a real number M > 0 such thatfor all n ∈ N,

xn ≥ M.

3. {xn} is said to be bounded when it is both bounded above and bounded below, i.e.,if there exists a real number M > 0 such that for all n ∈ N,

|xn| ≤ M.

Note that, the sequence {xn} is bounded if and only if the set {xn : n ∈ N} of its valueis a bounded subset of R.

Theorem 2.8 A convergent sequence of real numbers is bounded.

Proof. Suppose thatlim

n→∞xn = L

and ε = 1. Then there exists a natural number N such that for all n ≥ N ,

|xn − L| < 1.

By applying the triangle inequality (theorem 1.19 in chapter 1), we can obtain for n ≥ N

|xn| = |xn − L + L| ≤ |xn − L|+ |L| < 1 + |L| .

Now, if we setM := sup {|x1| , |x2| , · · · , |xN−1| , 1 + |L|} ,

then it follows that |xn| ≤ M for all n ∈ N.

Example 2.9 The sequence {xn} defined by

xn :=

{0 if n is odd1 if n is even

is bounded but has no limit. This example shows that the converse of theorem 2.8 doesnot hold.

2.2 Limit Theorems

In this section, we collect some miscellaneous theorems which are often useful inproving limits. Before starting, we will examine how the limit process interacts with thealgebraic operations of addition, substraction, multiplication and division of sequences.

Let X = {xn} and Y = {yn} are sequences of real numbers. Then we define :

18 Sequences

1. Sum of X and Y :X + Y = {xn + yn : n ∈ N} .

2. Difference of X and Y :

X − Y = {xn − yn : n ∈ N} .

3. Product of X and Y :XY = {xnyn : n ∈ N} .

4. Multiple of X by k ∈ R :

kX = {kxn : n ∈ N} .

5. Quotient of X and Y :X

Y=

{xn

yn

: n ∈ N}

with yn 6= 0 for all n ∈ N.

We now show that sequences obtained by applying these operations to convergentsequences give rise to new sequences whose limits can be predicted.

Theorem 2.10 Let {xn} and {yn} be sequences of real numbers that converges to x andy, respectively. Then

1. For k ∈ R, {kxn} converges to kx.

2. {xn + yn} converges to x + y.

3. {xnyn} converges to xy.

4. If {yn} is a sequence of nonzero numbers that converges to nonzero number y then{xn

yn

}converges to x

y.

Proof. Proof of 1. is very easy. So, we will prove remaining properties.

2. By hypothesis, for given ε > 0 there exists a natural number N1 such that ifn ≥ N1 then

|xn − x| < ε

2.

Similarly, there exists a natural number N2 such that if n ≥ N2 then

|yn − y| < ε

2.

Hence, if N = max {N1, N2}, it follows that if n ≥ N then

|(xn + yn)− (x + y)| ≤ |xn − x|+ |yn − y| < ε

2+

ε

2= ε.

Therefore,lim

n→∞(xn + yn) = x + y.


3. In order to prove this property, we will consider the following estimation :

|xnyn − xy| = |(xnyn − xny) + (xny − xy)|≤ |xn(yn − y)|+ |(xn − x)y|= |xn| |yn − y|+ |y| |xn − x| .

Since {xn} is a convergent sequence, according to theorem 2.8, there exists a realnumber M1 > 0 such that for all n ∈ N,

|xn| ≤ M1.

If we set M := max {M1, |y|} then we can obtain the following estimation

|xnyn − xy| ≤ M |yn − y|+ M |xn − x| .

From the convergence of {xn} and {yn}, we can say that if ε > 0 is given thenthere exist natural numbers N1 and N2 such that if n ≥ N1 and n ≥ N2 then

|xn − x| < ε

2Mand |yn − y| < ε

2M,

respectively. Now, by taking N = max {N1, N2}, we can infer that if n ≥ N then

|xnyn − xy| ≤ M |yn − y|+ M |xn − x| < Mε

2M+ M

ε

2M= ε.

Therefore,lim

n→∞xnyn = xy.

4. By 3., it is enough to show that

limn→∞

1

yn

=1

y.

Since {yn} converges, there exists a natural number N1 such that if n ≥ N1 then

|yn − y| < |y|2

.

From corollary 1.20,

−|y|2≤ − |yn − y| ≤ |yn| − |y|

for n ≥ N1, whence it follows that

|y|2

= |y| − |y|2

< |y| − |y − yn| ≤ |y − (y − yn)| = |yn|

for n ≥ N1. Therefore1

|yn|≤ 2

|y|for n ≥ N1 so we have the following estimation∣∣∣∣ 1

yn

− 1

y

∣∣∣∣ =

∣∣∣∣y − yn

yny

∣∣∣∣ =1

|yn| |y||y − yn| ≤

2

|y|2|y − yn| .

20 Sequences

Now, if ε > 0 is given then there exists a natural number N2 such that if n ≥ N2

then|y − yn| <

1

2ε |y|2 .

By taking N = max {N1, N2} then for n ≥ N∣∣∣∣ 1

yn

− 1

y

∣∣∣∣ ≤ 2

|y|2|y − yn| <

2

|y|2

(1

2ε |y|2

)= ε.

Therefore,

limn→∞

1

yn

=1

y

and by using 3., we can deduce that

limn→∞

xn

yn

=x

y.

Some of the results of theorem 2.10 can be extended, by Mathematical Induction,to a finite number of convergent sequences. For example, if A = {an}, B = {bn}, · · · ,Z = {zn} are convergent sequences of real numbers then their sum

A + B + · · ·+ Z = {an + bn · · · , zn}

is a convergent sequence and

limn→∞

(an + bn + · · ·+ zn) = limn→∞

an + limn→∞

bn + · · ·+ limn→∞

zn.

Also their product is a convergent sequence and

limn→∞

(anbn · · · zn) =(

limn→∞

an

) (lim

n→∞bn

)· · ·

(lim

n→∞zn

).

Moreover, if m ∈ N then Am is a convergent sequence and

limn→∞

(an)m =(

limn→∞

an

)m

.

Example 2.11 Since

lim1

n2= 0,

we can calculate the following :

limn→∞

2n2 − n

3n2 + 2= lim

n→∞

2− 1n

3 + 1n2

=2

3,

limn→∞

n + 3

n2 + 5n= lim

n→∞

1n

+ 3n2

1 + 5n

=0

1= 0

Theorem 2.12 If {xn} is a convergent sequence of real numbers and if xn ≥ 0 for alln ∈ N then

x = limn→∞

xn ≥ 0.


Proof. Suppose that the conclusion is not true, i.e., x < 0, then ε := −x is positive.Since {xn} converges to x, there exists a natural number N such that for all n ≥ N ,

x− ε < xn < x + ε.

In particular, we have xN < x + ε = x + (−x) = 0. This contradicts the hypothesis thatxn ≥ 0 for all n ∈ N. Therefore, this contradiction implies that

x = limn→∞

xn ≥ 0.

We now consider a useful result that is formally stronger than theorem 2.12.

Theorem 2.13 If {xn} and {yn} are convergent sequences of real numbers and if xn ≤yn for all n ∈ N then

limn→∞

xn ≤ limn→∞

yn.

Proof. Let us define zn := yn − xn then xn ≥ 0 for all n ∈ N. It follows from theorems2.10 and 2.12 that

0 ≤ limn→∞

zn = limn→∞

yn − limn→∞

xn.

Thereforelim

n→∞xn ≤ lim

n→∞yn.

Corollary 2.14 If {xn} is a convergent sequence and if a ≤ xn ≤ b for all n ∈ N then

a ≤ limn→∞

xn ≤ b.

The next result asserts that if a sequence {zn} is squeezed between two sequencesthat converges to the same limit, then it must also converge to this limit.

Theorem 2.15 (Squeeze theorem) Let {xn} and {yn} are convergent sequences ofreal numbers such that

limn→∞

xn = limn→∞

yn = L.

If {zn} be a sequence of real numbers such that xn ≤ zn ≤ yn for all n ∈ N then {zn}convergent and

limn→∞

zn = L.

Proof. From the convergence of {xn} and {yn}, for given ε > 0, there exists a naturalnumbers N1 and n2 such that if n ≥ N1 and n ≥ N2 then

|xn − L| < ε and |yn − L| < ε,

respectively. From the hypothesis, we can say that for all n ∈ N,

xn − L ≤ zn − L ≤ yn − L

it follows that|zn − L| ≤ max {|xn − L| , |yn − L|}.

Hence, by taking N := max {N1, N2}, we can deduce that

|zn − L| ≤ max {|xn − L| , |yn − L|} < ε.

22 Sequences

Example 2.16 Computelim

n→∞

n

10n.

Proof. Since n2 < 10n,

0 <n

10n<

n

n2=

1

n.

Thereforelim

n→∞

n

10n= 0 since lim

n→∞0 = 0 and lim

n→∞

1

n= 0.

Example 2.17 Compute (see Fig. 2.2)

limn→∞

n1n .

Proof. Put xn = n1n − 1 then for every n ∈ N

xn ≥ 0.

Since 1 + xn = n1n , we can say that n = (1 + xn)n. By the binomial theorem, if n ≥ 2,

we have

n = (1 + xn)n = 1 + nxn +1

2n(n− 1)(xn)2 + · · · ≥ 1

2n(n− 1)(xn)2,

whence it follows that(xn)2 ≤ 2

n− 1.

Since xn ≥ 0 for every n ∈ N,

0 ≤ xn ≤√

2

n− 1.

Applying squeeze theorem,

limn→∞

xn = 0 therefore limn→∞

n1n = 1.

2.3 Monotone Sequences

Until now, the main method available for showing that a sequence is convergent isto identify it as a subsequence or an algebraic combination of convergent sequences.However, when this cannot be done, we have to fall back on definition 2.3 in order toestablish the existence of limit. The use of this method has the noteworthy disadvantagethat we must already know (or at least suspect) the correct value of limit and we thenverify that our suspicion is correct.

There are many cases, however, where there is no obvious candidate for the limit of agiven sequence, even though a preliminary analysis has led to the belief that convergencedoes take place. In this section, we give some results which are deeper than those inthe preceding sections and which can be used to establish the convergence of a sequencewhen no particular element presents itself as the value of limit.

2.3 - Monotone Sequences 23

0 20 40 60 80 100 120 140 160 180 2001

1.05

1.1

1.15

1.2

1.25

1.3

1.35

1.4

1.45

Fig. 2.2 – First 200 values of sequence xn = n1n − 1. xn is getting close to 1 when n is

increasing.

Definition 2.18 Let {xn} be a sequence of real numbers.1. {xn} is increasing sequence if it satisfies the inequalities

x1 ≤ x2 ≤ · · · ≤ xn ≤ xn+1 ≤ · · · .

2. {xn} is decreasing sequence if it satisfies the inequalities

x1 ≥ x2 ≥ · · · ≥ xn ≥ xn+1 ≥ · · · .

3. {xn} is strictly increasing sequence if it satisfies the inequalities

x1 < x2 < · · · < xn < xn+1 < · · · .

4. {xn} is strictly decreasing sequence if it satisfies the inequalities

x1 > x2 > · · · > xn > xn+1 > · · · .

5. {xn} is (strictly) monotone if it is either (strictly) increasing or (strictly) decrea-sing.

Example 2.19 The following sequences are increasing

{an} = {n : n ∈ N} , {bn} = {3n : n ∈ N} , {cn} =

{(1 +

1

n

): n ∈ N

}.

The following sequences are decreasing

{dn} =

{1

n: n ∈ N

}, {en} = {−2n : n ∈ N} .

The following sequences are not monotone

{fn} = {(−1)n : n ∈ N} , {gn} = {cos nπ : n ∈ N} .

24 Sequences

Now, we will introduce an important theorem.

Theorem 2.20 (Monotone convergence theorem) A monotone sequence of realnumbers is convergent if and only if it is bounded. Further :

1. If {xn} is bounded increasing sequence then

limn→∞

xn = sup {xn : x ∈ N} .

2. If {xn} is bounded decreasing sequence then

limn→∞

xn = inf {xn : x ∈ N} .

3. Bounded monotone sequence is convergent.

Proof. We will prove 1. only. Proof of 2. is a homework.

Let {xn} be a bounded increasing sequence and set S = {xn : x ∈ N}. Since {xn} isbounded, there exists a real number M such that

xn ≤ M

for all n ∈ N. According to the completeness property of real number (see axiom 1.30),the supremum

x = sup {xn : x ∈ N}

exists in R.

In order to show that lim xn = sup {xn : x ∈ N} let ε > 0 be given. Then x− ε is notan upper bound of set S and hence there exists N ∈ N such that

x− ε < xN .

The fact that {xn} is increasing sequence implies that xN ≤ xn whenever n ≥ N , sothat for all n ≥ N ,

x− ε < xN ≤ xn ≤ x < x + ε.

Therefore we have|xn − x| < ε

for all n ≥ N . Therefore, we can conclude that

limn→∞

xn = sup {xn : x ∈ N} .

The monotone convergence theorem establishes the existence of the limit of a boun-ded monotone sequence. It also gives us a way of calculating the limit of the sequenceprovided we can evaluate the supremum (in case 1.) or the infimum (in case 2.). So-metimes it is difficult to evaluate this supremum or infimum, but once we know that itexists, it is often possible to evaluate the limit by other methods.

2.3 - Monotone Sequences 25

Example 2.21 (Recurrence formula) Let {yn} be defined inductively by

y1 = 3, yn+1 =yn

2+

3

yn

for n ≥ 1. Show that {yn} is convergent and

limn→∞

yn =√

6.

Proof. Since y1 = 3 > 0, yn > 0 for all n ∈ N and,

yn+1 − yn =yn

2+

3

yn

− yn =6− (yn)2

2yn

.

It is clear that yn is decreasing. So, in order to apply theorem 2.20, we now show, byinduction, that yn >

√6 for all n ∈ N.

The truth of this assertion can be verified for n = 1 since y1 = 3 >√

6. Now supposethat yk >

√6 for some k then

√6yk > 6 and

1

2>

3√6yk

implies1

2(yk −

√6) >

3√6yk

(yk −√

6).

So one can obtainyk

2−√

6

2>

3√6− 3

yk

.

Therefore,

yk+1 =yk

2+

3

yk

>

√6

2+

3√6

=√

6.

We have shown that the sequence yn is decreasing and bounded below by√

6. Itfollows from the theorem 2.20, yn is convergent sequence.

Unfortunately, in this case, it is not so easy to evaluate the lim yn by calcula-ting inf {yn : x ∈ N}. However, there is another way to evaluate. Let lim yn = L thenlim yn+1 = L also. By applying theorem 2.10, we can say

L =L

2+

3

L=⇒ L =

√6,−

√6.

Since yn > 0 for all n ∈ N, L 6= −√

6 implies

L = limn→∞

yn =√

6.

We end this section by introducing a sequence that converges to one of the mostimportant transcendental numbers in mathematics.

Example 2.22 (Euler’s number e) Let {xn} be a sequence of real numbers such thatfor all n ∈ N,

xn =

(1 +

1

n

)n

.

26 Sequences

We will show that this sequence is bounded and increasing ; hence it is convergent. Thelimit of this sequence is the famous Euler’s number e, whose approximate value is

e ≈ 2.718281828459045 · · · ,

which is taken as the base of the ‘natural’ logarithm, refer to Fig. 2.3.

Proof. If we apply the binomial theorem, we have

xn = nC01n + nC11

n−1 1

n+ · · ·+ nCk1

n−k

(1

n

)k

+ · · ·+ nCn

(1

n

)n

= 1 +n∑

k=1

nCk1n−k

(1

n

)k

:= 1 +n∑

k=1

yk.

Similarly,

xn+1 = 1 +n+1∑k=1

n+1Ck1n+1−k

(1

n + 1

)k

:= 1 +n+1∑k=1

zk.

Then for k = 1, 2, · · · ,

zk = n+1Ck1n+1−k

(1

n + 1

)k

=(n + 1)n(n− 1) · · · (n + 1− k + 1)

k!

(1

n + 1

)k

=n + 1

n + 1· n

n + 1· n− 1

n + 1· · · n + 1− k + 1

n + 1

1

k!

= 1 ·(

1− 1

n + 1

)·(

1− 2

n + 1

)· · ·

(1− k − 1

n + 1

)1

k!

≥ 1 ·(

1− 1

n

)·(

1− 2

n

)· · ·

(1− k − 1

n

)1

k!

=n

n· n− 1

n· n− 2

n + 1· · · n + 1− k

n + 1

1

k!

=n(n− 1)(n− 2) · · · (n + 1− k)

k!

(1

n

)k

= nCk1n−k

(1

n + 1

)k

= yk.

Therefore, xn ≤ xn+1 for all n ∈ N, so that {xn} is an increasing sequence. In orderto show {xn} is bounded, we will apply the following inequality (see exercise 1.1.11 ofmain textbook)

2n−1 ≤ n!

2.4 - Subsequences and the Cauchy criterion 27

for all n ∈ N. Then n−th term of {xn} is

xn =nC01n + nC11

n−1 1

n+ · · ·+ nCk1

n−k

(1

n

)k

+ · · ·+ nCn

(1

n

)n

=1 + n · 1

n+

n(n− 1)

2!

(1

n

)2

+ · · ·+ n(n− 1) · · · (n− k + 1)

k!

(1

n

)k

+ · · ·+(

1

n

)n

=1 + 1 +1

2!

(1− 1

n

)+ · · ·+ 1

n!

(1− 1

n

) (1− 2

n

)· · ·

(1− n− 1

n

)≤1 +

1

1!+

1

2!+ · · · 1

k!+ · · ·+ 1

n!

≤1 +1

20+

1

21+ · · · 1

2k−1+ · · ·+ 1

2n−1

=1 +1−

(12

)n

1− 12

< 1 +1

1− 12

= 3.

Hence, we deduce that {xn} is bounded sequence, so that {xn} converges by the mono-tone convergence theorem.

0 20 40 60 80 100 120 140 160 180 2002

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

Fig. 2.3 – First 200 values of sequence xn =(1 + 1

n

)n. xn is getting close to e when nis increasing.

2.4 Subsequences and the Cauchy criterion

In this section we will introduce the notion of a subsequence of a sequence of realnumbers. Informally, a subsequence of a sequence is a selection of terms from the givensequence such that the selected terms form a new sequence. Usually, subsequences arevery useful in establishing the convergence or the divergence of sequence. We will alsoprove the important existence theorem known as the Bolzano-Weierstrass theorem, whichwill be used to establish a number of significant results.

Definition 2.23 (Subsequence) Let {xn} be a sequence of real numbers and let n1 <n2 < · · · , nk < · · · be a strictly increasing sequence of natural numbers. Then {xnk

} :={xnk

}∞k=1 is called a subsequence of {xn}.

28 Sequences

Example 2.24 Let

{xn} =

{1

n: n ∈ N

}then the selection of even indexed terms produces the subsequence as follows ;

{xnk} :=

{1

nk

: k ∈ N}

=

{1

2k: k ∈ N

}where n1 = 2, n2 = 4, · · · , nk = 2k, · · · .

Subsequences of convergent sequences also converge to the same limit, as we nowshow :

Theorem 2.25 If a sequence {xn} of real numbers converges to a real number L if andonly if any subsequence {xnk

} of {xn} converges to L.

Proof. Let ε > 0 be given and let N ∈ N be such that if n ≥ N then

|xn − L| < ε.

Since n1 < n2 < · · · < nk < · · · is an increasing sequence of natural numbers, it can beproved (by induction) that nk ≥ k. Hence if k ≥ N , we also have nk ≥ k ≥ N so that

|xnk− L| < ε.

Therefore, the subsequence {xnk} converges to L.

Conversely, since {xn} is a subsequence of itself2 and any subsequence of {xn}converges to L, {xn} converges to L.

Corollary 2.26 Let {xn} be a sequence of real numbers1. If {xn} converges and there exists a subsequence which converges to L then {xn}

converges to L.2. If {xn} has two convergent subsequences whose limits are not equal then {xn}

diverges.3. If a subsequence of {xn} diverges then {xn} diverges.

Now, we will prove the important existence theorem known as the Bolzano-Weierstrass theorem : a bounded sequence of real numbers has a convergent subsequence.For that purpose, we will also prove the nested interval theorem.

Definition 2.27 We say that a sequence of intervals {In : n ∈ N} is nested if the follo-wing chain of inclusions holds

I1 ⊇ I2 ⊇ · · · ⊇ In ⊇ In+1 ⊇ · · · .

2By taking nk := k for k ∈ N.


Example 2.28 If for n ∈ N,

In :=

[0,

1

n

]then it is clear that In ⊇ In+1 for each n ∈ N so that this sequence of intervals is nested.In this case, the element 0 belongs to all In and the Archimedean property (theorem 1.32)can be used to show that 0 is the only such common point. We denote this by writing

∞⋂n=1

In = {0} .

Generally, a nested sequence of intervals need not have a common point. Let usconsider the following example.

Example 2.29 If for n ∈ N,

Jn :=

(0,

1

n

)then this sequence of intervals is nested, but there is no common point because for everygiven x > 0, there exists m ∈ N such that

x >1

m

so that x /∈ Jm. We denote this by writing

∞⋂n=1

Jn = φ.

It is an important property of R that every nested sequence of closed, boundedintervals does have a common point (see example 2.28). Notice that the completenessof R plays an essential role in establishing this property.

Theorem 2.30 (Nested intervals property) If In = [an, bn], n ∈ N, is a nestedsequence of closed bounded intervals then there exists a number x ∈ R such that x ∈ In

for all n ∈ N.

Theorem 2.31 If In = [an, bn], n ∈ N, is a nested sequence of closed bounded intervalssuch that the lengths bn − an of In satisfy

limn→∞

(bn − an) = 0

then the number x ∈ In for all n ∈ N is unique.

Proof. Since In ⊇ In+1, it is clear that an ≤ an+1 ≤ bn+1 ≤ bn. Let us define

S = {an : n ∈ N}

30 Sequences

then, since S 6= φ and an ≤ b1 for all n ∈ N, S is bounded above so that there exists asupremum of S. Let us denote this supremum as

x = sup S.

Moreover, {an} is an increasing sequence, by the monotone convergence theorem (theo-rem 2.20), we can say that

limn→∞

an = x.

Let us notice that in fact, it is essential to show that x ∈ In for all n ∈ N if and onlyif an ≤ x ≤ bn (refer to [2]). In order to show the uniqueness of x, let y ∈ In thenan ≤ y ≤ bn for all n ∈ N. Since

0 ≤ y − an ≤ bn − an

and lim(bn − an) = 0, by the squeeze theorem (theorem 2.15),

limn→∞

(y − an) = 0 implies x = limn→∞

an = y.

Therefore, we can conclude that x = y is the only point that belongs to In for everyn ∈ N.

We will now use the above theorem 2.31 to prove an important Bolzano-Weierstrasstheorem, which states that every bonded sequence has a convergent subsequence.

Theorem 2.32 (Bolzano-Weierstrass theorem) A bounded sequence of real num-bers has a convergent subsequence.

Proof. Let {xn} be a bounded sequence then for all n ∈ N, there exists a positive realnumber M such that

|xn| < M.

1. For all n ∈ N, we define an interval I0 = [a0, b0] satisfying

xn ∈ [−M, M ] = I0.

We now bisect I0 into two equal subintervals

I ′0 = [−M, 0] and I ′′0 = [0, M ].

2. One of these intervals must contain xn for infinitely many positive numbers n ∈ N.We denote this interval by I1 = [a1, b1].

3. We repeat this process with the interval I1, i.e., we bisect I1 into two equal subin-tervals I ′1 and I ′′1 . Notice that if I1 = I ′0 then

I ′1 =

[−M,−M

2

]and I ′′1 =

[−M

2, 0

].

4. Similarly with the previous case, one of these intervals I ′1 and I ′′1 must contain xn

for infinitely many positive numbers n ∈ N. We denote this interval by I2 = [a2, b2].


5. Continuing this process, we can obtain a nested sequence of interval {In} satisfying

I0 ⊇ I1 ⊇ I2 ⊇ · · ·

and a subsequence {xnk} of {xn} such that {xnk

} ∈ Ik for k ∈ N.6. Since the length of interval In is

(bn − an) =M

2n−1,

we can obtain

limn→∞

(length of interval In) = limn→∞

M

2n−1= lim

n→∞(bn − an) = 0.

Therefore, by theorem 2.31, there exists a unique common point x ∈ In for alln ∈ N.

7. Moreover, since {xnk} and x both belongs to Ik, we have

|xnk− x| < M

2k−1

whence it follows that the subsequence {xnk} of {xn} converges to x.

We now introduce the important notion of a Cauchy sequence in R. It will turn outthat a sequence in R is convergent if and only if it is a Cauchy sequence. It is importantfor us to have a condition implying the convergence of a sequence that does not require usto know the value of the limit of sequence in advance (and is not restricted to monotonesequences).

Definition 2.33 (Cauchy sequence) A sequence {xn} of real numbers is said to bea Cauchy sequence if for every ε > 0 there exists a natural number N such that forall natural numbers n,m ≥ N , the terms xn and xm satisfy

|xn − xm| < ε.

Example 2.34 The sequence

{xn} =

{1

n: n ∈ N

}is a Cauchy sequence.

Proof. If ε > 0 is given, we choose a natural number N such that 1N

< ε. Then ifm,n ≥ N and n > m (m > n case is similar), we have

|xn − xm| =∣∣∣∣ 1

m− 1

n

∣∣∣∣ =n−m

nm<

n

nm=

1

m<

1

N< ε.

Therefore, we conclude that {xn} is a Cauchy sequence.

Example 2.35 The sequence {xn} = {1 + (−1)n : n ∈ N} is not a Cauchy sequence.

32 Sequences

One of our purpose is to show that the Cauchy sequences are precisely the convergentsequences. First, we will prove that a convergent sequence is a Cauchy sequence.

Lemma 2.36 If {xn} is a convergent sequence of real numbers, then {xn} is a Cauchysequence.

Proof. Assume that {xn} converges to x. Then for given ε > 0 there exists a naturalnumber N such that if n ≥ N then

|xn − x| < ε

2.

Thus, if m, n ≥ N then

|xn − xm| = |xn − x + x− xm| ≤ |xn − x|+ |xm − x| < ε

2+

ε

2= ε.

Therefore, {xn} is a Cauchy sequence.

Next, we will show the following result.

Lemma 2.37 A Cauchy sequence of real numbers is bounded.

Proof. Let {xn} be a Cauchy sequence and let ε := 1. Then there exists a naturalnumber N such that if n ≥ N then

|xn − xN | < 1.

Hence, by the triangle inequality, we have |xn| ≤ |xN |+ 1 for all n ≥ N . If we set

M := max {|x1|, |x2|, · · · , |xN−1|, |xN |+ 1} ,

then it follows that |xn| ≤ M for all n ∈ N. Therefore, {xn} is abounded sequence.

Now, we present the important Cauchy convergence criterion :

Theorem 2.38 (Cauchy convergence criterion) A sequence of real numbers isconvergent if and only if it is a Cauchy sequence.

Proof. As we seen, in Lemma 2.36, that a convergent sequence of real numbers is aCauchy sequence. Conversely, Let {xn} be a Cauchy sequence. We will show that it isconvergent to some real number.

1. We observe from Lemma 2.37 that {xn} is bounded.2. Therefore, by the Bolzano-Weierstrass theorem (theorem 2.32), there is a subse-

quence {xnk} of {xn} that converges to some real number L.

3. Since {xn} is a Cauchy sequence, given ε > 0 there exists a natural number N1

such that if n, m ≤ N1 then|xn − xm| <

ε

2.


4. Since {xnk} converges to L, there exists a natural number N2 such that if k ≤ N2

then|xnk

− L| < ε

2.

5. Let N = max {N1, N2}. If k ≤ N , since nk ≤ k ≤ N ,

|xk − L| ≤ |xk − xnk|+ |xnk

− L| < ε

2+

ε

2= ε.

We infer thatlim

n→∞xn = L.

Therefore, {xn} is convergent.

We will now give an example of application of the Cauchy criterion

Example 2.39 Let {xn} be a sequence defined by

x1 = 1, x2 = 2, xn =1

2(xn−1 + xn−2) for n > 2.

Then {xn} is a convergent sequence.

Proof. Since, |x2−x1| = 1 and |x3−x2| =1

2, it can be shown by mathematical induction

that for all n ∈ N,1 ≤ xn ≤ 2.

Some calculations shows that this sequence is not monotone. However, since the termsare formed by averaging, by mathematical induction, it is readily seen that for all n ∈ N,

|xn − xn−1| =1

2n−2.

Thus, if m > n, we may employ the triangle inequality in order to obtain

|xm − xn| ≤ |xm − xm−1|+ |xm−1 − xm−2|+ · · ·+ |xn+1 − xn|

=1

2m−2+

1

2m−3+ · · ·+ 1

2n+

1

2n−1

=1

2n−1

(1

2m−n−1+

1

2m−n−2+ · · ·+ 1

22+

1

2+ 1

)<

1

2n−1× 2 =

1

2n−2.

Therefore, given ε > 0, if N is chosen so large that

1

2N<

ε

4

and if m ≤ n ≤ N then it follows that

|xm − xn| <1

2n−2≤ 1

2N−2=

4

2N< ε.

Therefore, the sequence {xn} is a Cauchy sequence in R. By the Cauchy criterion (theo-rem 2.38), we infer that {xn} converges.

34 Sequences

Remark 2.40 In order to evaluate the limit L of above sequence {xn}, we might firstpass to the limit in the rule of definition

xn =1

2(xn−1 + xn−2)

to conclude that L must satisfy the relation

L =1

2(L + L),

which is true, but not informative. Hence, we must try something else.

Since {xn} converges to L, so does the subsequence {x2n+1} with odd indices. Bymathematical induction, it can be shown that

x2n+1 = 1 +1

2+

1

23+ · · ·+ 1

22n−1= 1 +

2

3

(1− 1

4n

).

It follows from this that

L = limn→∞

xn = limn→∞

x2n+1 = 1 +2

3=

5

3.

The following contractive sequence written in the exercise of main textbook is animportant sequence in analysis. So, we will finish this section with following :

Definition 2.41 (Contractive sequence) We say that a sequence {xn} of real num-bers is contractive if there exists a constant α, 0 < α < 1 such that

|xn+2 − xn+1| ≤ α|xn+1 − xn|

for all n ∈ N. The number α is called the constant of the contractive sequence.

Theorem 2.42 Every contractive sequence is a Cauchy sequence, and therefore isconvergent.

Proof. This is a homework.

Corollary 2.43 If {xn} is a contractive sequence with constant α, 0 < α < 1, and if

L = limn→∞

xn,

then

|xn − L| ≤ αn−1

1− α|x2 − x1|.

|xn − L| ≤ α

1− α|xn − xn−1|.

2.5 - Upper and Lower Limits of Bounded and Unbounded Sequences 35

Proof. Since {xn} is contractive sequence, if m > n then

|xm − xn| ≤αn−1

1− α|x2 − x1|.

Therefore, if we let m −→∞ then

|xn − L| ≤ αn−1

1− α|x2 − x1|.

Next, recall that if m > n then by triangle inequality

|xm − xn| = |xm − xm−1 + xm−1 − xm−2 − · · · − xn+1 + xn+1 − xn|≤ |xm − xm−1|+ |xm−1 − xm−2|+ · · ·+ |xn+1 − xn|.

Since it is readily established, using mathematical induction, that

|xn+k − xn+k−1| ≤ αk|xn − xn−1|,

we infer that

|xm − xn| ≤ (αm−n + · · ·+ α2 + α)|xn − xn−1| ≤α

1− α|xn − xn−1|.

We now let m −→∞ in this inequality in order to obtain assertion 2.

2.5 Upper and Lower Limits of Bounded and Unbounded Se-quences

The generalized limits lim sup xn and lim inf xn are defined for arbitrary (not neces-sary convergent) sequences {xn}. In this section, we will define limit superior and limitinferior for bounded or unbounded sequences {xn}. If {xn} is a bonded sequence, theBolzano-Weierstrass theorem assures us that {xn} has a convergent subsequence. Thelimit superior and limit inferior of {xn} is the maximum and minimum value obtainableas the limit of a convergent subsequence of {xn}, respectively.

For certain purposes it is convenient to define what is meant for a sequence {xn} ofreal numbers to diverges to ±∞ (or tend to ±∞).

Definition 2.44 (Divergent sequences) Let {xn} be a sequence of real numbers.

1. We say that {xn} diverges to infinity (or tends to infinity) if for every M ∈ R,there exists a natural number N such that if n ≥ N then

xn > M

and writelim

n→∞xn = +∞.

36 Sequences

2. We say that {xn} diverges to minus infinity (or tends to minus infinity) iffor every M ∈ R, there exists a natural number N such that if n ≥ N then

xn < M

and writelim

n→∞xn = −∞.

3. We say that {xn} is properly divergent in case we have either

limn→∞

xn = +∞ or limn→∞

xn = −∞.

We should realize that we are using the symbols +∞ and −∞ purely as a convenientnotation in the above expressions. Results that have been proved in earlier sections forconventional limits lim xn = L (for L ∈ R) may not remain true when lim xn = ±∞.

Theorem 2.45 Let {xn} and {yn} be two sequences of real numbers such that

limn→∞

xn = +∞ and limn→∞

yn > 0

thenlim

n→∞xnyn = +∞.

Proof. Let M be a positive real number. Since, lim yn > 0, choose a positive number Lsuch that

0 < L < limn→∞

yn.

Then there exists a natural number N1 such that, if n ≤ N1 then

yn > L.

Since lim xn = +∞, there exists a natural number N2 such that, if n ≤ N2 then

xn >M

L.

Let N = max {N1, N2} then if n ≥ N then

xnyn >M

LL = M

Therefore, we infer that lim xnyn = +∞.

Monotone sequences are particularly simple in regard to their convergence. We haveseen in the monotone convergence theorem that a monotone sequence is convergent ifand only if it is bounded. The next theorem is a reformulation of that result.

Theorem 2.46 A monotone sequence of real numbers is properly divergent if and onlyif it is unbounded.


1. If {xn} is an unbounded increasing sequence then

limn→∞

xn = +∞.

2. If {xn} is an unbounded decreasing sequence then

limn→∞

xn = −∞.

Proof. Suppose that {xn} is an unbounded increasing sequence. Then for any M ∈ R,there exists N ∈ N such that

xN > M.

But since {xn} is increasing, for all n ≥ N , we have

xn > M.

Since M is arbitrary, it follows that lim xn = +∞. Remaining part can be proved in asimilar fashion.

The following comparison theorem is frequently used in showing that a sequence isproperly divergent.

Theorem 2.47 Let {xn} and {yn} be two sequences of real numbers and suppose thatfor all n ∈ N,

xn ≤ yn.

Then the followings are holds :

If limn→∞

xn = +∞ then limn→∞

yn = +∞.

If limn→∞

yn = −∞ then limn→∞

xn = −∞.

Proof. Let lim xn = +∞ and M ∈ R is given. Then there exists a natural number Nsuch that, if n ≥ N then

M < xn ≤ yn︸︷︷︸by hypothesis

implies M < yn.

Since M is arbitrary, it follows that lim yn = +∞. The proof of 2. is similar.

Let us notice that since it is sometimes difficult to establish an inequality such asxn ≤ yn, the following limit comparison theorem is often more convenient to use.

Theorem 2.48 (Limit comparison theorem) Let {xn} and {yn} be two sequencesof positive real numbers and suppose that for some positive real number L > 0, we have

limn→∞

xn

yn

= L.

Thenlim

n→∞xn = +∞ if and only if lim

n→∞yn = +∞.

38 Sequences

The following theorem is also useful for obtaining the limit of sequences.

Theorem 2.49 Let {xn} be a sequence of real numbers such that x > 0 for all n ∈ N.Then

limn→∞

xn = +∞ if and only if limn→∞

1

xn

= 0.

Proof. Let lim xn = +∞ and for given ε > 0, set M = 1ε∈ R. Then there exists a

natural number N such that, if n ≥ N then

xn > M =1

εimplies

1

xn

< ε.

Therefore ∣∣∣∣ 1

xn

− 0

∣∣∣∣ < ε implies limn→∞

1

xn

= 0.

Conversely, let us assume that lim 1xn

= 0 and let ε = 1M

for a positive number M ∈ R.Then there exists a atural number N ∈ N such that, if n ≥ N then∣∣∣∣ 1

xn

− 0

∣∣∣∣ < ε =1

M.

Since xn > 0 for all n ∈ N,

0 <1

xn

< M.

Therefore, xn > M for all n ∈ N, we have

limn→∞

xn = +∞.

Now, let us consider the limit superior and limit inferior of an arbitrary sequence.

Definition 2.50 Let {xn} be a sequence of real numbers1. Let Ak = sup {xk, xk+1, · · · } = sup {xn : n ≥ k}. Then L is the limit superior of{xn} if

L := limk→∞

Ak = limk→∞

sup xk.

2. Let Bk = inf {xk, xk+1, · · · } = inf {xn : n ≥ k}. Then L is the limit inferior of{xn} if

L := limk→∞

Bk = limk→∞

inf xk.

The notations limxn and limxn are also used for lim sup xn and lim inf xn, respectively.

Theorem 2.51 Let {xn} be a sequence of real numbers then

limxn = infn

supk≥n

{xk} and limxn = supn

infk≥n

{xk} .

Proof. See the theorem 2.28 and exercise 3 of section 2.5 of main textbook.


Example 2.52 Compute the limit superior and limit inferior of sequence

{xn} =

{(−1)n +

1

n: n ∈ N

}.

Proof. Let us define a set Ak = {xn : n ≥ k, n ∈ N}. If k is an even number then k + 1is an odd number and so on. Then

Ak =

{1 +

1

k,−1 +

1

k + 1, 1 +

1

k + 2,−1 +

1

k + 3· · ·

}implies

sup Ak = 1 +1

kand inf Ak = −1.

Similarly, if k is an odd number then

Ak =

{−1 +

1

k, 1 +

1

k + 1,−1 +

1

k + 2, 1 +

1

k + 3· · ·

}implies

sup Ak = 1 +1

k + 1and inf Ak = −1.

Therefore,limxn = 1 and limxn = −1.

Example 2.53 Compute the limit superior and limit inferior of sequence

{xn} =

{1

n: n ∈ N

}.

Proof. Let us define a set Ak = {xn : n ≥ k, n ∈ N}. Then

Ak =

{1

k,

1

k + 1,

1

k + 2,

1

k + 3· · ·

}implies

sup Ak =1

kand inf Ak = 0.

Therefore,limxn = limxn = 0.

Above example says that if a sequence is convergent, its limit superior and limitinferior are same.

Theorem 2.54 Let {xn} be a bonded sequence of real numbers. If

limn→∞

xn = L if and only if L = limxn = limxn.

40 Sequences

Proof. LetL = lim

n→∞xn.

Then for ε > 0 given, there exists a natural number N such that, if n ≥ N then

|xn − L| < ε

2.

Therefore, if n ≥ N then

L− ε

2< An = sup {xn, xn+1, · · · } ≤ L +

ε

2

impliesL− ε

2< lim

n→∞An ≤ L +

ε

2.

Since lim An = limxn,

L− ε < L− ε

2< limxn ≤ L +

ε

2< L + ε implies

∣∣limxn − L∣∣ < ε.

Therefore, L = limxn. Similarly, one can show that L = limxn.

Conversely, let us assume that L = limxn = limxn. Then, since

L = limxn = limn→∞

sup {xn, xn+1, · · · } ,

for ε > 0 given, there exists a natural number N1 such that, if n ≥ N1 then

|sup {xn, xn+1, · · · } − L| < ε implies xn < L + ε.

Similarly, sinceL = limxn = lim

n→∞inf {xn, xn+1, · · · } ,

for ε > 0 given, there exists a natural number N2 such that, if n ≥ N2 then

|inf {xn, xn+1, · · · } − L| < ε implies L− ε < xn.

Let N = max {N1, N2}. If n ≥ N then

L− ε < xn < L + ε implies |xn − L| < ε.

Therefore,lim

n→∞xn = L.

Corollary 2.55 Let {xn} be a sequence of real numbers. If

limn→∞

xn = ∞ if and only if limxn = limxn = ∞.

Proof. See the theorem 2.30 of main textbook.



1. Prove that(a) Let {an} and {bn} be sequences such that {an} is bounded and lim bn = 0.

Thenlim

n→∞anbn = 0.

(b) Give an example of sequences {an} and {bn} such that lim bn = 0 but

limn→∞

anbn 6= 0.

2. Let {an} and {bn} be sequences of real numbers and x ∈ R. If for some k > 0 andevery natural number n,

|xn − x| < k|an| and limn→∞

an = 0

thenlim

n→∞xn = x.

3. Establish the convergence or the divergence of the following sequences.

(a) xn =3− 2n

1 + n.

(b) xn =(−1)nn

2n− 1.

(c) xn =n2 − 2

n + 1.

(d) xn =1− n

2n.

(e) xn =n!

2n.

(f) xn =n2

2n.

4. Prove that every contractive sequence (see definition 2.41) is a Cauchy sequence.5. Prove the remaining part 2. of theorem 2.20 (theorem 2.13 of main textbook).6. Prove that if the subsequences {x2n} and {x2n−1} of {xn} converges to a real

number x then {xn} converges.7. Calculate the limit superior and limit inferior of following sequences.

(a) {xn} = {1 + (−1)n : n ∈ N}.

(b) {yn} =

{1

2, 1,

1

4,1

3,1

6,1

5,1

8,1

7, · · ·

}.

(c) {zn} ={n2(−1 + (−1)n) : n ∈ N

}.

(d) {tn} ={

n sinnπ

2: n ∈ N

}.

42 Limits of Functions


Mathematical analysis is generally understood to refer to that area of mathematicsin which systematic use is made of various limiting concepts : the limit of a sequence ofreal numbers. In this chapter, we will encounter the notion of the limit of function.

3.1 Limits of Functions

The intuitive idea of the function f having a limit L at the point a is that the valuesf(x) are close to L when x is close to (but different from) a. But it is necessary to havea technical way of working with the idea of close to and this is accomplished in the ε−δdefinition given in this section.

In order for the idea of the limit of a function f at a point a to be meaningful, it isnecessary that f be defined at the point close to a. It is need not be defined at the pointa, but it should be defined at enough points close to a to make the study interesting.These are the reasons for the following definitions :

Definition 3.1 (Neighborhood (revisited)) Let a ∈ R and ε > 0.1. The ε−neighborhood of a is the set

Nε(a) := {x ∈ R : |x− a| < ε} = {x ∈ R : a− ε < x < a + ε} .

2. D is called the neighborhood of a if there exists an ε−neighborhood Nε(a) suchthat

Nε(a) ⊂ D.

3. The ε−deleted neighborhood of a is the set

N∗ε (a) := {x ∈ R : 0 < |x− a| < ε} = {x ∈ R : a− ε < x < a + ε} − {a} .

Example 3.2 Let us consider the following examples of neighborhood :1. Let I := {x : 0 < x < 1} and a ∈ I. Let ε = min {a, 1− a} then Nε(a) is an

ε−neighborhood of a. Moreover, for arbitrary x ∈ Nε(a),

0 ≤ a− ε < x < a + ε ≤ 1 implies x ∈ I.

It means that Nε(a) is contained in I. Thus I is neighborhood of a.2. Let I := {x : 0 ≤ x ≤ 1} then for any ε > 0, Nε(0) contains points not in I, and

so Nε(0) is not contained in I. For example, the number x = − ε2

is in Nε(0) butnot in I.

3. Let a be a real number. For given ε > 0, if x ∈ Nε(a) then x = a by theorem 1.23.

Definition 3.3 Let D ⊆ R. A point x is an accumulation point or cluster point(or limit point) of D if for every δ−neighborhood Nδ(x) of x contains at least one pointof D distinct from a, i.e.,

(x− δ, x + δ) ∩ (D − {x}) 6= Ø.

3.1 - Limits of Functions 43

Let us notice that the point x may or may not be a member of D, but even if it isin D, it is ignored when deciding whether it is an accumulation point of D or not, sincewe explicitly require that there be points in Nδ(x)∩D distinct from x in order for x tobe an accumulation point of D. For the sake of simplicity, we set the domain D be anon-empty subset of R.

Example 3.4 Let us consider the following examples of accumulation point :

1. The set S =

{1

n: n ∈ N

}has only the point 0 as an accumulation point. None of

the points in S is a cluster point of S.

2. The set S = {0} ∪{

1

n: n ∈ N

}has only the point 0 as an accumulation point.

3. For intervals (0, 1), (0, 1], [0, 1) and [0, 1], every point of the closed interval [0, 1]is an accumulation point of them.

Theorem 3.5 A number a ∈ R is an accumulation point of a subset D ⊆ R if and onlyif there exists a sequence {an} in D such that for all n ∈ N

limn→∞

an = a and an 6= a.

Proof. If a is an accumulation point of D then for any n ∈ N, the 1n−neighborhood

N 1n(a) contains at least one point an in D distinct from a. Then

an ∈ A, an 6= a and |an − a| < 1

nimplies lim

n→∞an = a.

Conversely, if there exists a sequence {an} in D−{a} with limn→∞

an = a then for anyε > 0, there exists N ∈ N such that if n ≥ N then

an ∈ Nε(a).

Therefore, for n ≥ N , Nε(a) contains the points an such that

an ∈ D and an 6= a.

It means that a is an accumulation point of D.

We now state the precise definition of the limit of a function f at a point a. It isimportant to note that in this definition, it is immaterial whether f is defined at a ornot. In any case, we exclude a from consideration in the determination of the limit.

Definition 3.6 (Limit of function) Let D ⊆ R and a be an accumulation point ofD. A function f : D −→ R, a real number L is said to be a limit of f at a if for givenε > 0, there exists a δ(ε) > 0 such that if x ∈ D and 0 < |x− a| < δ(ε) then

|f(x)− L| < ε.

If the limit of f at a does not exists, we say that f diverges at a.


1. Let us notice that the notation δ(ε) is used to emphasize that the choice of δdepends on the value of ε. However, it is often convenient to write δ instead ofδ(ε). For the sake of simplicity, we will use δ instead of δ(ε).

2. If L is a limit of f at a then we also say that f converges to L at a. We often write

limx→a

f(x) = L.

3. Sometimes, the symbolism

f(x) −→ L as x −→ a

is used in order to indicate the intuitive idea that the f has limit L at a.

The following theorem indicates that the value of L of the limit of function is uniquelydetermined. This uniqueness is not part of the definition of limit, but must be deduced.

Theorem 3.7 (Uniqueness of limits) Let f : D −→ R be a function and if a is anaccumulation point of D then f can have only one limit at a.

Proof. Let L1 and L2 are both limits of f at a. Then for given ε > 0, there exists δ1 > 0such that if x ∈ D and 0 < |x− a| < δ1 then

|f(x)− L1| <ε

2.

Also there exists δ2 > 0 such that if x ∈ D and 0 < |x− a| < δ2 then

|f(x)− L2| <ε

2.

Now, let δ = min {δ1, δ2} then if a ∈ D and 0 < |x− a| < δ,

|L1 − L2| ≤ |L1 − f(x)|+ |f(x)− L2| <ε

2+

ε

2= ε.

Since ε > 0 is arbitrary, L1 = L2.

The definition of limit can be described in terms of neighborhoods, refer to Fig. 3.1.We observe that because

Nδ(a) = {x : |x− a| < δ} ,

the inequality 0 < |x−a| < δ is equivalent to saying that x 6= a and x ∈ Nδ(a). Similarly,the inequality |f(x)− L| < ε is equivalent to saying that f(x) ∈ Nε(L). In this way wecan obtain the following result. The proof is left to reader.

Theorem 3.8 Let f : D −→ R be a function and a be an accumulation point of D.Then the following statements are equivalent :

1. limx→a

f(x) = L.

2. Given any Nε(L), there exists a Nδ(a) such that if x 6= a is any point in Nδ(a)∩Dthen f(x) ∈ Nε(L).


L

Nε(L)

Nδ(a)

a

given

there exists

x

y

Fig. 3.1 – The limit of f at a.

Example 3.9 Show thatlimx→a

b = b.

Proof. Let f(x) := b for all x ∈ R then if ε > 0 is given, we let δ = ε (in fact, anystrictly positive δ will serve the purpose, e.g., δ = 1). Then if 0 < |x− a| < δ then

|f(x)− b| = |b− b| = 0 < ε.

Since ε > 0 is arbitrary, we conclude that

limx→a

b = b.


x = a.

Proof. Let f(x) := x for all x ∈ R then if ε > 0 is given, we let δ = ε. Then if0 < |x− a| < δ then

|f(x)− a| = |x− a| < δ = ε.

Since ε > 0 is arbitrary, we deduce that

limx→a

x = a.


x2 = a2.

Proof. Let f(x) := x2 for all x ∈ R. We want to make the difference as

|f(x)− a2| = |x2 − a2| < ε


for a preassigned ε > 0 by taking x sufficiently close to a. To do so, we note thatx2 − a2 = (x + a)(x− a). Moreover if |x− a| < M then3

|x| ≤ |a|+ M so that |x + a| ≤ |x|+ |a| ≤ 2|a|+ M.

Therefore, if |x− a| < M , we have

|f(x)− a2| = |x2 − a2| = |x + a||x− a| ≤ (2|a|+ M)|x− a|. (3.1)

The last term of above inequality will be less than ε provided we take

|x− a| < ε

2|a|+ M.

Consequently, if we choose

δ := min

{M,

ε

2|a|+ M

},

then if 0 < |x − a| < δ, it follow first that |x − a| < M so that (3.1) is valid, andtherefore,

|f(x)− a2| = |x2 − a2| = |x + a||x− a| ≤ (2|a|+ M)ε

2|a|+ M< ε.

Since we have a way of choosing δ > 0 for an arbitrary choice of ε > 0, we infer that

limx→a

f(x) = limx→a

x2 = a2.

Example 3.12 Show that if a > 0,

limx→a

1

x=

1

a.

Proof. Let f(x) := 1x

for all x ∈ R. We want to make the difference as∣∣∣∣f(x)− 1

a

∣∣∣∣ =

∣∣∣∣1x − 1

a

∣∣∣∣ < ε

for a preassigned ε > 0 by taking x sufficiently close to a. To do so, we note that forx > 0, ∣∣∣∣1x − 1

a

∣∣∣∣ =

∣∣∣∣ 1

ax(a− x)

∣∣∣∣ =1

ax|x− a|.

It is useful to get an upper bound for the term 1ax

that holds in some neighborhood ofa. In particular, if |x− a| < 1

2a then 1

2a < x < 3

2a, so that

0 <1

ax<

2

a2for |x− a| < 1

2a.

Therefore, for these values of x we have∣∣∣∣1x − 1

a

∣∣∣∣ =1

ax|x− a| ≤ 2

a2|x− a|. (3.2)

3In our main textbook, M = 1 is used.


The last term of above inequality will be less than ε provided we take |x − a| < 12a2ε.


δ := min

{1

2a,

1

2a2ε

},

then if 0 < |x − a| < δ, it follow first that |x − a| < 12a so that (3.2) is valid, and

therefore, ∣∣∣∣f(x)− 1

a

∣∣∣∣ =1

ax|x− a| ≤ 2

a2

1

2a2ε < ε.

Since we have a way of choosing δ > 0 for an arbitrary choice of ε > 0, we infer that

limx→a

f(x) = limx→a

1

x=

1

a.

There are times when a function f may not posses a limit at a point a, yet a limitdoes exist when the function is restricted to an interval on one side of the accumulationpoint a. For example, the following signum function sgn defined by (see Fig. 3.2)

sgn(x) :=

+1 for x > 0

0 for x = 0−1 for x < 0

has no limit at a = 0. However, if we restrict the sgn(x) to the interval (0,∞), theresulting function has a limit of 1 at a = 0. Similarly, if we restrict the sgn(x) to theinterval (−∞, 0), the resulting function has a limit of −1 at a = 0. These are elementaryexamples of right-hand and left-hand limits at a = 0.

x

y

0

1

-1

f (x)=sgn(x)

Fig. 3.2 – The signum function f(x) = sgn(x).

Definition 3.13 (Right-hand and Left-hand limits) Let f : D −→ R be a func-tion.

1. If a is an accumulation point of D ∩ (a,∞), then we say that L ∈ R is a right-hand limit of f at a if given any ε > 0, there exists a δ > 0 such that for allx ∈ D with 0 < x− a < δ,

|f(x)− L| < ε.

In this case, we write

limx→a+

f(x) = L or f(a+) = L.


2. If a is an accumulation point of D ∩ (−∞, a), then we say that L ∈ R is a left-hand limit of f at a if given any ε > 0, there exists a δ > 0 such that for allx ∈ D with 0 < a− x < δ,

|f(x)− L| < ε.

In this case, we write

limx→a−

f(x) = L or f(a−) = L.

Let us notice that the limits limx→a+

f(x) and limx→a−

f(x) are called one-sided limits of f

at a. It is possible that neither one-sided limit may exists. Also, one of them may existwithout the other existing. Similarly, as is the case for f(x) = sgn(x) at x = 0, theymay both exist and be different.

The following result relates the notion of the limit of function to one-sided limits.

Theorem 3.14 Let f : D −→ R be a function and a be an accumulation point ofD ∩ (a,∞) and D ∩ (−∞, a). Then

limx→a

f(x) = L if and only if limx→a+

f(x) = L = limx→a−

f(x).

Proof. Suppose thatlimx→a

f(x) = L.

For given ε > 0, there exists δ > 0 such that if 0 < |x− a| < δ then

|f(x)− L| < ε.

If 0 < x− a < δ then 0 < |x− a| < δ ; so

|f(x)− L| < ε implies limx→a+

f(x) = L.

Similarly, one can obtainlim

x→a−f(x) = L.

Conversely, suppose that

limx→a+

f(x) = L = limx→a−

f(x).

For given ε > 0, there exists δ1 > 0 such that if 0 < x− a < δ1 then

|f(x)− L| < ε.

Moreover, there exists δ2 > 0 such that if 0 < a− x < δ2 then

|f(x)− L| < ε.

Let δ = min {δ1, δ2} then if 0 < |x− a| < δ either 0 < x− a < δ1 or 0 < a− x < δ2 sothat

|f(x)− L| < ε implies limx→a

f(x) = L.


Example 3.15 For x 6= 0, let us consider the function

f(x) := |x|+ x

|x|.

Thenlim

x→0+f(x) = 1 and lim

x→0−f(x) = −1.

Example 3.16 Calculate the right-hand and left-hand limit of the function f(x) atx = 1 (see Fig. 3.3)

f(x) :=

2x + 1 for x > 1

x

2for x ≤ 1

Proof. For given ε > 0 there exists δ = ε2

> 0 such that if 1 < x < 1+ δ then |x−1| < δso that

|f(x)− 3| = |(2x + 1)− 3| = 2|x− 1| < 2δ = ε.

Therefore, we infer thatlim

x→1+f(x) = 3.

For given ε > 0 there exists δ = 2ε > 0 such that if 0 < 1− x < δ then

|x− 1| < δ

so that

|f(x)− 3| =∣∣∣∣x2 − 1

2

∣∣∣∣ =1

2|x− 1| < 1

2δ = ε.

Therefore, we infer that

limx→1−

f(x) =1

2.

0 1

x

y

0.5

3

y = f (x)

Fig. 3.3 – Graph of f(x) in example 3.16.


3.2 Some Properties of Limits of Functions

In this section, we shall obtain some results that are useful in calculating limitsof functions. These results are parallel to the limit theorems established in section 2.2for sequences. In fact, in most cases these results can be proved by using results fromthe previous chapter. Alternatively, some results in this section can be proved by usingtypical ε − δ arguments that are very similar to the ones employed in the previouschapter.

We will start this section with the following important formulation of limit of afunction is in terms of limits of sequences. This characterization permits the theory ofChapter 2 to be applied to the study of limits of functions.

Theorem 3.17 (Sequential criterion) Let f : D −→ R be a function and a be anaccumulation point of D. Then the following are equivalent.

1. limx→a

f(x) = L.

2. For every sequence {xn} in D that converges to a such that xn 6= a for all n ∈ N,

limn→∞

f(xn) = L.

Proof. Assume thatlimx→a

f(x) = L.

For given ε > 0 there exists δ > 0 such that if 0 < |x− a| < δ then

|f(x)− L| < ε.

Since {xn} converges to a, for given δ > 0, there exists N ∈ N such that if n ≥ N then

|xn − a| < δ.

But for each xn, xn 6= a, we have 0 < |xn − a| so that ; if n ≥ N then 0 < |xn − a| < δ,we have

|f(xn)− a| < ε.

Therefore, {f(xn)} converges to L.

Conversely, in order to apply the contrapositive argument, assume that

limx→a

f(x) 6= L.

Then there exists Nε0(L) such that no matter what Nδ(a) we pick, there exists at leastone number xδ with 0 < |xδ − a| < δ and xδ 6= a such that

|f(xδ)− L| ≥ ε0.

Hence for every n ∈ N, there exists N 1n(a) contains a number xn ∈ D such that

0 < |xn − a| < 1

nbut |f(xn)− L| ≥ ε0 for all n ∈ N.

3.2 - Some Properties of Limits of Functions 51

So, we conclude that the sequence {xn} in D that converges to a such that xn 6= a, butthe sequence {f(xn)} does not converges to L. This is contradiction. So, we have

limx→a

f(x) = L.

Sometimes, it is important to be able to show that a certain number is not the limitof a function at a point or that the function does not have a limit at a point. Thefollowing result is a consequence of theorem 3.17.

Theorem 3.18 (Divergence criterion) Let f : D −→ R be a function and a be anaccumulation point of D. Then the following are equivalent.

1. limx→a

f(x) 6= L.

2. There exists a sequence {xn} in D with xn 6= a for all n ∈ N such that

limn→∞

xn = a but limn→∞

f(xn) 6= L.

Example 3.19 Let

f(x) := sin1

x

then limx→0

f(x) does not exist in R (See Fig. 3.4).

−0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x−axis

y−ax

is

1

π

1

3π

1

2π


Proof. Now, we recall from Elementary Calculus that for integer n,

sin x =

{0 if x = nπ1 if x = π

2+ 2nπ.

For n ∈ N, let xn := 1nπ

then

limn→∞

xn = 0 and f(xn) = sin nπ = 0 so that limn→∞

f(xn) = 0.


On the other hand, for n ∈ N, let yn :=(π

2+ 2nπ

)−1

then

limn→∞

yn = 0 and f(yn) = sin(π

2+ 2nπ

)= 1 so that lim

n→∞f(yn) = 1.

Therefore,lim

n→∞f(xn) 6= lim

n→∞f(yn).

This implies that

limx→0

f(x) = limx→0

sin1

x

does not exist.

Definition 3.20 (Bounded function) Let f : D −→ R be a function and let a be anaccumulation point of D. We say that f is bounded on a neighborhood of a if thereexists a Nδ(a) and a constant M > 0 such that for all a ∈ D ∩Nδ(a),

|f(x)| ≤ M.

Theorem 3.21 Let f : D −→ R be a function and a be an accumulation point of D. If

limx→a

f(x) = L

then f is bounded on some neighborhood of a.

Proof. Sincelimx→a

f(x) = L,

for ε = 1, there exists δ > 0 such that if 0 < |x− a| < δ then

|f(x)− L| < 1 implies |f(x)| − |L| ≤ |f(x)− L| < 1.

Therefore, if x ∈ D ∩Nδ(a) and x 6= a, then |f(x)| < L + 1. Now let

M =

{|L|+ 1 if a /∈ D

sup {|f(a)|, |L|+ 1} if a ∈ D.

It follows that if x ∈ D ∩Nδ(a) then

|f(x)| ≤ M.

This shows that f is bounded on some neighborhood Nδ(a) of a.

Theorem 3.22 Let f : D −→ R be a function and a be an accumulation point of D. If

limx→a

f(x) = L > 0

then there exists a neighborhood Nδ(a) such that f(x) > 0 for all x ∈ D ∩ Nδ(a) andx 6= a.


Proof. Sincelimx→a

f(x) = L > 0,

suppose that ε = L2

> 0. Then there exists δ > 0 such that if 0 < |x− a| < δ and x ∈ Dthen

|f(x)− L| < L

2implies − L

2< f(x)− L <

L

2.

Therefore it follows that if 0 < |x− a| < δ and x ∈ D then

f(x) >L

2> 0.

The next definition is similar to the definition for sums, differences, products, andquotients of sequences given in section 2.2.

Definition 3.23 Let f : D −→ R be a function and g : D −→ R be functions. Wedefine

1. Sum f + g :(f + g)(x) = f(x) + g(x).

2. Difference f − g :(f − g)(x) = f(x)− g(x).

3. Product fg :fg(x) = f(x)g(x).

4. Multiple kf for k ∈ R :(kf)(x) = kf(x).

5. Quotient fg

: (f

g

)(x) =

f(x)

g(x)

with g(x) 6= 0 for all x ∈ D.

Theorem 3.24 Let f : D −→ R and g : D −→ R be functions and a be an accumulationpoint of D. Further, let k ∈ R. If

limx→a

f(x) = L and limx→a

g(x) = M

then :1. lim

x→a(f + g)(x) = L + M .

2. limx→a

kf(x) = kL.

3. limx→a

(f − g)(x) = L−M .

4. limx→a

(fg)(x) = LM .

5. limx→a

(f

g

)(x) =

L

Mwhere M 6= 0.


Proof. The proof of this theorem is very similar to that of theorem 2.10. Notice thatone can prove this theorem by using Sequential criterion (theorem 3.17).

1. For given ε > 0, there exists a δ1 > 0 such that if x ∈ D and 0 < |x− a| < δ1 then

|f(x)− L| < ε

2.

Moreover, there exists a δ2 > 0 such that if x ∈ D and 0 < |x− a| < δ2 then

|g(x)−M | < ε

2.

Let us take δ = min {δ1, δ2}. If x ∈ D and 0 < |x− a| < δ then

|f(x) + g(x)− (L + M)| ≤ |f(x)− L|+ |g(x)−M | < ε

2+

ε

2= ε.

Therefore,limx→a

(f + g)(x) = L + M.

2. If k = 0 then this property holds so let us assume that k 6= 0 case. For given ε > 0,there exists δ > 0 such that if x ∈ D and 0 < |x− a| < δ then

|f(x)− L| < ε

|k|.

Analogously, if x ∈ D and 0 < |x− a| < δ then

|kf(x)− kL| = |k||f(x)− L| < |k| ε

|k|= ε.

Therefore,limx→a

kf(x) = kL.

3. By combining 1. and 2., we can prove it.

4. In order to prove this property, we will consider the following estimation :

|f(x)g(x)− LM | = |f(x)g(x)− Lg(x) + Lg(x)− LM |≤ |f(x)− L| |g(x)|+ |L| |g(x)−M | .

Since g has limit M , according to theorem 3.21, g is bounded on some neighbo-rhood of a so that ; there exists δ1 > 0 such that if x ∈ D and 0 < |x − a| < δ1

then|g(x)| < |M |+ 1.

By hypothesis, for given ε > 0 there exists δ2 > 0 such that if x ∈ D and0 < |x− a| < δ2 then

|f(x)− L| < ε

2(|M |+ 1).

And there exists δ3 > 0 such that if x ∈ D and 0 < |x− a| < δ3 then

|g(x)−M | < ε

2|L|.


Let us take δ = min {δ1, δ2, δ3}. If x ∈ D and 0 < |x− a| < δ then

|f(x)g(x)− LM | ≤ |f(x)− L| |g(x)|+ |L| |g(x)−M |

<ε

2(|M |+ 1)(|M |+ 1) + |L| ε

2|L|<

ε

2+

ε

2= ε.

Therefore,limx→a

(fg)(x) = LM.

5. By 4., it is enough to show that

limx→a

1

g(x)=

1

M.

Let us notice that ∣∣∣∣ 1

g(x)− 1

M

∣∣∣∣ =|M − g(x)||g(x)M |

=|g(x)−M ||g(x)||M |

.

Since g has limit M , there exists δ1 > 0 such that if x ∈ D and 0 < |x − a| < δ1

then||g(x)| − |M || ≤ |g(x)−M | < |M |

2implies

|M |2

< |g(x)|.

Therefore, if x ∈ D and 0 < |x− a| < δ1 then

1

|g(x)|<

2

|M |.

Moreover, for given ε > 0, there exists δ2 > 0 such that if x ∈ D and 0 < |x−a| < δ2

then|g(x)−M | < |M |2

2ε.

Let us take δ = min {δ1, δ2}. If x ∈ D and 0 < |x− a| < δ then∣∣∣∣ 1

g(x)− 1

M

∣∣∣∣ =|g(x)−M ||g(x)||M |

<2

|M |2|M |2

2ε = ε.

Therefore, by combining 3., we infer that

limx→a

(f

g

)(x) =

L

M

where M 6= 0.

Remark 3.25 Let us note that in part 6., the additional assumption that

limx→a

g(x) = M 6= 0

is made. If this assumption is not satisfied, then the limit

limx→a

f(x)

g(x)

may or may not exist. But even if this limit does exist, we cannot use property 6. oftheorem 3.24 to evaluate it.


Remark 3.26 fk : D −→ R for k = 1, 2, · · · , n be functions and a be an accumulationpoint of D. If

limx→a

fk(x) = Lk for k = 1, 2, · · · , n

then it follows from theorem 3.24 by an Mathematical induction argument that

limx→a

(f1 + f2 + · · ·+ fn)(x) = L1 + L2 + · · ·+ Ln

andlimx→a

(f1f2 · · · fn)(x) = L1L2 · · ·Ln.

In particular, we deduce that if limx→a

f(x) = L then for n ∈ N

limx→a

{f(x)}n = Ln.

Example 3.27 If a 6= 0 then

limx→a

1

x=

limx→a

1

limx→a

x=

1

a.

Example 3.28 Let f(x) = x2 − 4 and g(x) = 3x − 6 for x ∈ R then we cannot applyproperty 6. of theorem 3.24 to evaluate

limx→2

f(x)

g(x)= lim

x→2

x2 − 4

3x− 6

becauselimx→2

(3x− 6) = 0.

However, if x 6= 2, then it follows that

x2 − 4

3x− 6=

(x + 2)(x− 2)

3(x− 2)=

x + 2

3.

Therefore, we have

limx→2

f(x)

g(x)= lim

x→2

x2 − 4

3x− 6= lim

x→2

x + 2

3=

4

3.

Example 3.29 limx→0

f(x) = limx→0

1

xdoes not exist in R.

Proof. Since limx→0

x = 0, we cannot apply property 6. of theorem 3.24 to evaluate limx→0

f(x).In order to show that lim

x→0f(x) does not exist, we will apply theorem 3.18. For that

purpose, let xn = 1n

thenlim

n→∞xn = 0.

But, the sequence {f(xn)} = {n : n ∈ N} does not bounded, i.e., limn→∞

f(xn) does notexist. Therefore, by theorem 3.18, lim

x→0f(x) does not exist in R.

The next theorem is a direct analogue of theorem 2.13.


Theorem 3.30 fk : D −→ R for k = 1, 2, · · · , n be functions and a be an accumulationpoint of D. If for x ∈ D and x 6= a,

f(x) ≤ g(x) and there exists limx→a

f(x) and limx→a

g(x)

thenlimx→a

f(x) ≤ limx→a

g(x).

Proof. We will apply the contrapositive argument for proving theorem. Let us assumethat

limx→a

f(x)− limx→a

g(x) = limx→a

{f(x)− g(x)}︸︷︷︸by 3. of theorem 3.24

= k > 0.

For ε = k2

> 0, there exists δ > 0 such that if x ∈ D and 0 < |x− a| < δ then

|f(x)− g(x)− k| < ε =k

2so that − k

2< f(x)− g(x)− k <

ε

2.

Therefore, if x ∈ D and 0 < |x− a| < δ then

f(x)− g(x) >k

2> 0.

This is contradiction. So, we have

limx→a

f(x) ≤ limx→a

g(x).

We now state an analogue of the squeeze theorem 2.15. We leave its proof to thereader.

Theorem 3.31 (Squeeze theorem) Let f, g, h : D −→ R be functions and a be anaccumulation point of D. If for x ∈ D and x 6= a,

f(x) ≤ g(x) ≤ h(x) and limx→a

f(x) = L = limx→a

h(x)

thenlimx→a

g(x) = L.

Theorem 3.32 Let f, g : D −→ R be functions and a be an accumulation point of D.If for x ∈ D and x 6= a,

g(x) is bounded and limx→a

f(x) = 0

thenlimx→a

f(x)g(x) = 0.

Example 3.33 Prove that

limx→0

x sin1

x= 0.


Proof. Letf(x) = x sin

1

xfor x 6= 0. Since for all x ∈ R, −1 ≤ sin x ≤ 1, we have the following inequality

−|x| ≤ f(x) = x sin1

x≤ |x|

for all x ∈ R, x 6= 0. Since limx→0

x = 0, it follows from the squeeze theorem 3.31 that

limx→0

x sin1

x= 0.

For a graph, see Fig. 3.5.

−0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4−0.25

−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

0.25

x−axis

y−ax

is

1

π

1

2π

1

3π


Let us consider the functionf(x) =

1

x2

for x 6= 0, refer to Fig. 3.6. f(x) is not bounded on a neighborhood of 0, so it cannothave a limit in the sense of definition 3.6. While the symbols ∞(= +∞) and −∞ do notrepresent real numbers, it is sometimes useful to be able to say that f(x) approaches(or tends) to ∞ as x −→ 0. This use of ±∞ will not cause any difficulties, provided weexercise caution and never interpret ∞ or −∞ as being real numbers.

Definition 3.34 Let f : D −→ R be a function and a be an accumulation point of D.1. We say that f approaches to infinity (or tends to infinity) as x −→ a if for

every M ∈ R there exists δ = δ(M) > 0 such that for all x ∈ D with 0 < |x−a| < δthen

f(x) > M

and writelimx→a

f(x) = ∞ (or +∞).


−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2

0

2

4

6

8

10

x0

y

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−5

−4

−3

−2

−1

0

1

0

x

y

Fig. 3.6 – Graph of f(x) = 1/x2 (left) and f(x) = log |x| (right) for x 6= 0.

2. We say that f approaches to minus infinity (or tends to minus infinity)as x −→ a if for every M ∈ R there exists δ = δ(M) > 0 such that for all x ∈ Dwith 0 < |x− a| < δ then

f(x) < M

and writelimx→a

f(x) = −∞.

Example 3.35 Prove that (see Fig. 3.6)

limx→0

1

x2= ∞.

Proof. For every M ∈ R, there exists δ =1√M

such that if 0 < |x− 0| < δ then

x2 <1

Mso that

1

x2> M.

Example 3.36 Prove that (see Fig. 3.6)

limx→0

log |x| = −∞.

Proof. For every M ∈ R, there exists δ = eM such that if 0 < |x− 0| < δ then

|x| < eM so that log |x| < M.

Similarly with the definition 3.13, it will be useful to consider one-sided infinite limitsas

limx→a+

f(x) = ∞, limx→a−

f(x) = ∞, limx→a+

f(x) = −∞ and limx→a−

f(x) = −∞.


Example 3.37 For x 6= 0,

f(x) =1

xdoes not tend to either ∞ or −∞ as x −→ 0. In fact

limx→0+

f(x) = ∞ and limx→0−

f(x) = −∞.

It is also desirable to define the notion of the limit of a function as x −→ ∞. Thedefinition as x −→ −∞ is similar.

Definition 3.38 Let f : D −→ R be a function.1. We say that L ∈ R is a limit of f as x −→ ∞ if given any ε > 0 there exists M

such that for any x > M , then

|f(x)− L| < ε

and writelim

x→∞f(x) = L.

2. We say that f approaches to infinity (or tends to infinity) as x −→ ∞ ifgiven any M ∈ R there exists K ∈ R such that for any x > K, then

f(x) > M

and writelim

x→∞f(x) = ∞.

3. Similarly, we can define

limx→−∞

f(x) = L, limx→−∞

f(x) = ∞, limx→∞

f(x) = −∞ and limx→−∞

f(x) = −∞.

Example 3.39 Compute

limx→∞

(− 1

2x + 3

)

Proof. For given ε > 0 there exists M =1

2ε> 0 such that if x > M then∣∣∣∣− 1

2x + 3− 0

∣∣∣∣ =1

2x + 3<

1

2x<

1

2M< ε.

Therefore,

limx→∞

(− 1

2x + 3

)= 0.

Example 3.40 Computelim

x→∞

√x

Proof. For any M > 0, there exists K = M2 > 0 such that if x > K then√

x >√

K = M implies limx→∞

√x = ∞.



1. Show that the following limit does not exist

limx→0

x

|x|.

2. Evaluate the following limits, or show that they do not exist.

(a) limx→0−

2|x|x

.

(b) limx→2+

x2 − 4

|x− 2|.

(c) limx→0+

∣∣∣∣sin 1

x

∣∣∣∣ .

(d) limx→0+

x1x .

(e) limx→0+

4x[x]

2x + |x|.

(f) limx→0−

4x[x]

2x + |x|.

3. A function f : R −→ R satisfies

f(x + y) = f(x) + f(y)

for all x, y ∈ R and there exists limx→0

f(x). Show that the following holds.

(a) limx→0

f(x) = 0.

(b) For all a ∈ R, there exists limx→a

f(x).

4. Prove the following by using squeeze theorem 3.31 (or theorem 3.16 of main text-book).(a) lim

x→0sin x = 0.

(b) limx→0

cos x = 1.

(c) limx→0

sin x

x= 1.

(d) limx→0

cos x− 1

x= 0.

62 Continuous Functions


We now begin the study of the most important class of functions that arises inanalysis : the class of continuous functions.

4.1 Continuous Functions

In this section, we will define what it means to say that a function is continuous at apoint, or on a set. This notion of continuity is one of the central concepts of mathematicalanalysis, and it will be used in almost all of the following material. For convenience, weset the domain D be a non-empty subset of R.

Definition 4.1 (Continuous function) Let f : D −→ R be a function and let a ∈ D.We say that f is continuous at a if, given any number ε > 0 there exists δ > 0 suchthat if x ∈ D satisfying |x− a| < δ then

|f(x)− f(a)| < ε.

If f is continuous on every point of D, then we say that f is continuous on D. If ffails to be continuous at a, then we say that f is discontinuous at a.

As with the definition of limit, the definition of continuity at a point can be formu-lated in terms of neighborhoods. This is done in the next result.

Theorem 4.2 A function f : D −→ R is continuous at a point a ∈ D if and only ifgiven any ε > 0 there exists δ > 0 such that if x ∈ D ∩Nδ(a) then

f(x) ∈ Nε(f(a)).

Let us notice that if a is an accumulation point of D, then a comparison of definitions3.6 and 4.1 show that f is continuous at a if and only if

limx→a

f(x) = f(a).

Thus, if a is an accumulation point of D, then three conditions must hold for f to becontinuous at a :

1. f must be defined at a (so that f(a) makes sense)

2. the limit of f at a must exist in R (so that limx→a

f(x) makes sense)

3. these two values must be equal (so that limx→a

f(x) = f(a)).

Example 4.3 Constant function, f(x) = x and f(x) = x2 are continuous functions,refer to examples 4.1, 4.2 and 4.3 of main textbook.

4.1 - Continuous Functions 63

Example 4.4 Let f : R −→ R be a function defined as

f(x) =

x sin

1

xif x 6= 0

0 if x = 0

then f is continuous at x = 0.

Proof. For any ε > 0, there exists δ = ε > 0 such that if |x− 0| < δ, x ∈ R then

|f(x)− f(0)| =∣∣∣∣x sin

1

x

∣∣∣∣ ≤ |x| < δ = ε.

Therefore, combining the result in example 3.33, f is continuous at x = 0.

A slight modification of the proof of theorem 3.17 for limits yields the followingsequential version of continuity at a point.

Theorem 4.5 (Sequential criterion for continuity) A function f : D −→ R iscontinuous at the point a ∈ D if and only if for every sequence {xn} in D that convergesto a, the sequence {f(xn)} converges to f(a).

The following discontinuity criterion is a consequence of the last theorem.

Theorem 4.6 (Discontinuity criterion) A function f : D −→ R is discontinuous atthe point a ∈ D if and only if for every sequence {xn} in D that converges to a, but thesequence {f(xn)} does not converges to f(a).

Example 4.7 Let f : R −→ R be a function defined by

f(x) :=

{0 if x is rationalx if x is irrational.

We claim that f is continuous at x = 0 and is discontinuous except x = 0.

Proof. For given ε > 0, there exists δ = ε > 0 such that if |x− 0| < δ then

|f(x)− f(0)| = |f(x)− 0| = |f(x)| ≤ |x| < δ = ε.

Therefore,limx→0

f(x) = 0,

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

Now, assume that x 6= 0 then since |x| > 0, there exists n0 ∈ N such that

|x| > 1

n0

.


For each n ∈ N, let us define

xn = x− 1

n0 + n.

First, let us assume that x is irrational number. Since xn is irrational number, thereexists a rational number yn such that

xn < yn < x.

Since limn→∞

xn = x, by squeeze theorem

limn→∞

yn = x.

Therefore by the sequential criterion for continuity (theorem 4.5),

limn→∞

f(yn) = 0 6= x = f(x). (4.1)

Next, let us assume that x is rational number. Since xn is rational number, thereexists an irrational number yn such that

xn < yn < x.

Since limn→∞

xn = x, by squeeze theorem

limn→∞

yn = x.

Therefore by the sequential criterion for continuity (theorem 4.5),

limn→∞

f(yn) = x 6= 0 = f(x). (4.2)

Since every real number if either rational or irrational, by (4.1) and (4.2), f is disconti-nuous at x 6= 0.

Example 4.8 Let f : R −→ R be Dirichlet’s discontinuous function4 defined by

f(x) :=

{1 if x is rational0 if x is irrational.

We claim that f is not continuous at any point of R.

Proof. If a be a rational number, let {xn} be a sequence of irrational numbers suchthat5

limn→∞

xn = a.

Since f(xn) = 0 for all n ∈ N, we have

limn→∞

f(xn) = 0 6= 1 = f(a).

4This function was introduced in 1829 by P. G. L. Dirichlet.5The density theorem 1.35 assures us that such a sequence does exist.


On the other hand, if a be an irrational number, let {xn} be a sequence of rationalnumbers such that6

limn→∞

xn = a.

Since f(xn) = 0 for all n ∈ N, we have

limn→∞

f(xn) = 1 6= 0 = f(a).

Since every real number if either rational or irrational, we deuce that f is not continuousat any point in R.

Example 4.9 Let h : (0, 1) −→ R be Thomae’s function7 defined by

h(x) :=

1

nif x =

m

nfor m, n ∈ N and gcd(m, n) = 1

0 if x is irrational.

We claim that h is not continuous at every irrational number in (0, 1), and is disconti-nuous at every rational number in (0, 1), refer to Fig. 4.1.

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

Fig. 4.1 – Graph of h(x). Left : largest n = 10. Right : largest n = 20.

Proof. If c ∈ (0, 1) be a rational number, let {xn} be a sequence of irrational numbersin (0, 1) such that

limn→∞

xn = c.

Since h(xn) = 0 for n ∈ N,lim

n→∞h(xn) = 0.

But c is rational, h(c) > 0. Therefore,

limn→∞

h(xn) 6= h(c).

6Similarly with the previous situation, the density theorem 1.35 assures us that such a sequencedoes exist.

7This function was introduced in 1875 by K. J. Thomae.


Hence h is discontinuous at rational number c ∈ (0, 1).

On the other hand, if d ∈ (0, 1) be an irrational number and ε > 0 then by theArchimedean property (theorem 1.32), there exists n0 ∈ N such that

1

n0

< ε.

Let us define a set of rational numbers

Qn0 ={m

n: 1 ≤ n ≤ n0, 0 < m < n, gcd(m, n) = 1

}=

{1

2,1

3,2

3,1

4,3

4,1

5,2

5, · · · ,

1

n0

, · · · ,n0 − 1

n0

}.

Then there are only a finite number of rationals in Qn0 . For d /∈ Qn0 , if we set δ > 0such that

δ = min {|d− x| : x ∈ Qn0}then the neighborhood (d−δ, d+δ) contains no rational numbers with denominator lessthan n0. It means that n > n0 since, a rational number

m

n∈ (0, 1) satisfies∣∣∣m

n− d

∣∣∣ < δ.

It follows that for |x− d| < δ, x ∈ (0, 1), we have

|h(x)− h(d)| = |h(x)| ≤ 1

n0

< ε.

Thus h is continuous at the irrational number d ∈ (0, 1).

The next result is similar to theorem 3.24, from which it follows.

Theorem 4.10 Let f, g : D −→ R be functions and k ∈ R. Suppose that a ∈ D andthat f and g are continuous at a. Then

1. f + g, f − g, fg and kf are continuous at a.

2. If g(x) 6= 0 for all x ∈ D then the quotientf

gis continuous at a.

The next result is an immediate consequence of theorem 4.10, applied to every pointof D. This is an important result.

Theorem 4.11 Let f, g : D −→ R be continuous on D and k ∈ R.1. f + g, f − g, fg and kf are continuous on D.

2. If g(x) 6= 0 for all x ∈ D then the quotientf

gis continuous on D.

Example 4.12 A polynomial function

p(x) = a0 + a1x + a2x2 + · · ·+ an−1x

n−1 + anxn

is continuous on R.


Example 4.13 f(x) = sin x is continuous on R.

Proof. For all x, y, z ∈ R, we have

| sin z| ≤ |z|, | cos z| ≤ 1 and sin x− sin y = 2 sinx− y

2cos

x + y

2.

Hence for a ∈ R, we have

0 ≤ | sin x− sin a| =∣∣∣∣2 sin

x− a

2cos

x + a

2

∣∣∣∣ ≤ 2

∣∣∣∣x− a

2

∣∣∣∣ = |x− a|.

Since limx→a

|x− a| = 0, by squeeze theorem,

limx→a

sin x = sin a.

Therefore f(x) = sin x is continuous at x = a. Since a ∈ R is arbitrary, it follows thatf is continuous on R.

Example 4.14 g(x) = x−1 is continuous on R− {0}.

In the next result, we will show that the composition of continuous functions is alsoa continuous function.

Theorem 4.15 Let A, b ⊆ R and let f : A −→ R and g : B −→ R be functions suchthat f(A) ⊆ B. If f and g are continuous at a ∈ A and b = f(a) ∈ B, respectively, thenthe composition

g ◦ f : A −→ R

is continuous at a.

Proof. Since g is continuous at b = f(a) ∈ B, for given ε > 0, there exists δ1 > 0 suchthat if y ∈ B and |y − f(a)| < δ1 then

|g(y)− g(f(a))| < ε.

Since f is continuous at a ∈ A, with this δ1 > 0, there exists δ > 0 such that if x ∈ Aand |x− a| < δ then

|f(x)− f(a)| < δ1 implies |g(f(x))− g(f(a))| < ε.

It follows that g ◦ f is continuous at a.

Example 4.16 Since f(x) = x−1 is continuous at every point on R − {0} and g(x) =sin x is continuous on R,

(g ◦ f)(x) = sin1

x

is continuous at every point on R− {0}.


4.2 Properties of Continuous Functions

Functions that are continuous on intervals have a number of important propertiesthat are not possessed by general continuous functions. In this section, we will establishsome results that are of considerable importance and that will be applied later.

Definition 4.17 A function f : D −→ R is said to be bounded on D if there exists aconstant M > 0 such that

|f(x)| ≤ M

for all x ∈ D. On the other hand, F is said to be unbounded on D if given M > 0,there exists a point x ∈ D such that

|f(x)| > M.

Example 4.18 f(x) =√

x is bounded on [0, 2].

Example 4.19 Let us consider the function f defined on the interval (0, 1) by

f(x) =1

x.

For any M > 0, we can take the point x0 =1

M + 1∈ (0, 1) such that

|f(x0)| =1

x0

= M + 1 > M.

Therefore, f is not bounded on (0, 1).

Example 4.19 shows that continuous functions need net be bounded however, we’llshow that continuous functions on a certain type of interval are necessarily bounded.

Theorem 4.20 (Boundedness theorem) Let f : [a, b] −→ R be continuous on [a, b].Then f is bounded on [a, b].

Proof. Suppose that f is unbounded on [a, b]. Then for any n ∈ N there exists a numberxn ∈ [a, b] such that

|f(xn)| > n.

Since [a, b] is bounded interval, {xn} is a bounded sequence. Therefore, the Bolzano-Weierstrass theorem 2.32 implies that there is a subsequence {xnk

} such that

limk→∞

xxk= c.

Since a ≤ xnk≤ b, by theorem 3.30, a ≤ c ≤ b so that c ∈ [a, b]. Since f is continuous

at climk→∞

f(xxk) = f(c).

We then conclude from theorem 2.8 that the convergent sequence {f(xxk)} must be

bounded. But this is contradiction since

limk→∞

|f(xxk)| ≥ lim

k→∞|nk| = ∞.

Therefore, f is bounded on [a, b].

4.2 - Properties of Continuous Functions 69

Remark 4.21 To show that each hypothesis of the boundedness theorem is need, we canconstruct examples that show the conclusion fails if any one of the hypotheses is relaxed.

1. The interval must be bounded. For example, the function f(x) := x for x ∈ [0,∞)is continuous but unbounded on [0,∞).

2. The interval must be closed. For example, the function f(x) :=1

xfor x ∈ (0, 1] is

continuous but unbounded on (0, 1].3. The function must be continuous. For example, the function

f(x) :=

1

xfor x ∈ (0, 1]

1 for x = 0

is discontinuous and unbounded on [0, 1].

Now, we will prove that a continuous function on a closed bounded interval mustattain a maximum and a minimum value. This is very important.

Definition 4.22 (Maximum & Minimum) Let f : D −→ R be a function.1. We say that f has an absolute maximum on D if there is a point x∗ such that

f(x∗) ≥ f(x) for all x ∈ D.

In this case, x∗ is called an absolute maximum point for f on D if it exist.2. We say that f has an absolute minimum on D if there is a point x∗ such that

f(x∗) ≤ f(x) for all x ∈ D.

In this case, x∗ is called an absolute minimum point for f on D if it exist.

Example 4.23 f(x) = x2 defined on [−1, 1] has two points x = ±1 giving the absolutemaximum and the single point x = 0 yielding its absolute minimum (see Fig. 4.2).

Example 4.24 g(x) := x−1 has an absolute maximum but no absolute minimum on[1,∞) (see Fig. 4.2).

Theorem 4.25 (Maximum-Minimum theorem) Let f : [a, b] −→ R be a conti-nuous function on [a, b]. Then f has an absolute maximum and an absolute minimumon [a, b], i.e., there exists p, q ∈ [a, b] such that

f(p) ≤ f(x) ≤ f(q).

Proof. For convenience, we define I = [a, b] and

f(I) := {f(x) : x ∈ [a, b]}


1 2 3 4 5

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x−axis

y−ax

is

(a) Graph of f(x)

−1 −0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

x−axis

y−ax

is

(b) Graph of g(x)

By theorem 4.20, f is bounded therefore, there exists s∗ and s∗ such that

s∗ = sup f(I) and s∗ = inf f(I).

Since s∗ = sup f(I), if n ∈ N, then the number s∗ − 1

nis not an upper bound of the set

f(I). Consequently there exists a number xn ∈ I such that

s∗ − 1

n< f(xn) ≤ s∗ for all n ∈ N. (4.3)

Since I is bounded, by theorem 2.8, {xn} is bounded. Therefore, the Bolzano-Weierstrasstheorem 2.32, there exists a subsequence {xnk

} such that

limk→∞

xxk= q.

Since a ≤ xnk≤ b, by theorem 3.30, a ≤ q ≤ b so that q ∈ I. Therefore, f is continuous

at q so thatlimk→∞

f(xxk) = f(q).

Since it follows from (4.3) that

s∗ − 1

nk

< f(xnk) ≤ s∗ for all k ∈ N,

we conclude from the squeeze theorem 3.31 that

limk→∞

f(xxk) = s∗.

Therefore we have

f(q) = limk→∞

f(xxk) = s∗ = sup {f(x) : x ∈ [a, b]} .

We conclude that q is an absolute maximum point of f on [a, b]. The proof of theexistence of p is left to the reader.

The next result assures us that a continuous function on an interval takes on (atleast once) any number that lies between two of its values.

4.2 - Properties of Continuous Functions 71

Theorem 4.26 (Bolzano’s intermediate value theorem) Let f : [a, b] −→ R be acontinuous function on [a, b] and f(a) < f(b). If r ∈ R satisfies f(a) < r < f(b), thenthere exists a point c ∈ (a, b) such that

f(c) = r.

Proof. Let A = {x ∈ [a, b] : f(x) < r}. Since a ∈ A, A 6= φ. Moreover, A is boundedabove (by b), there exists a supremum

c := sup A.

Since b is an upper bound of A, c ≤ b.

Moreover, since c is an upper bound of A, for n ∈ N there exists {xn} such that

c− 1

n< xn ≤ c.

Since, by squeeze theorem, lim xn = c and xn ∈ A,

f(xn) < r.

By hypothesis, since f is continuous at c,

f(c) = limn→∞

f(xn) ≤ r.

Now, let us assume that f(c) < r. Let

ε :=1

2(r − f(c)) > 0

then since f is continuous at c, there exists δ > 0 such that if |x− c| < δ, x ∈ [a, b] then

|f(x)− f(c)| < ε implies f(c)− ε < f(x) < f(c) + ε.

Notice that (c, c + δ) ∩ (c, b] 6= φ since f(c) < r and c 6= b. So, for (c, c + δ) ∩ (c, b],

f(x) < f(c) + ε = f(c) +1

2(r − f(c)) =

1

2(r + f(c)) <

1

2(r + r) = r.

It means that x ∈ A and x > c. This contradicts to c = sup A. Therefore, f(c) = r.

As obvious property of intervals is that if two points a, b ∈ I with a < b then anypoint t satisfies a < t < b also belongs to I, refer to theorem 1.25. In other words, ifa, b ∈ I then [a, b] is contained in I.

Theorem 4.27 (Characterization theorem) If I is a subset of R that contains atleast two points and has the property :

if x, y ∈ I and x < y then [x, y] ⊆ I,

then I is an interval.


Proof. See [3, theorem 2.5.1].

The next theorem summarizes the main results of this part. It states that the image ofa closed bounded interval under a continuous function is also al closed bounded interval.The endpoints of the image interval are the absolute minimum and absolute maximumvalues of the function and the statement that all values between the absolute minimumand absolute maximum values belong to the image is a way of describing Bolzano’sintermediate value theorem.

Theorem 4.28 Let I be a closed bounded interval and let f : I −→ R be continuous onI. Then the set

f(I) := {f(x) : x ∈ I}

is a closed bounded interval.

The next theorem extends above result to general intervals. However, it should benoticed that although the continuous image of an interval is shown to be an interval, itis not true that the image interval necessarily has the same form as the domain interval.

Example 4.29 Letf(x) = sin x

for x ∈ R. Then f is continuous on R. It is easy to see that for given open intervalI = (0, 2π), the continuous image of an interval f(I) = [−1, 1] an open interval.

Theorem 4.30 (Preservation of intervals theorem) Let I be an interval and letf : I −→ R be continuous on I. Then the set f(I) is an interval.

Proof. Let s, t ∈ f(I) with s < t ; then there exists points a, b ∈ I (a 6= b) such that

f(a) = s and f(b) = t.

Further, it follows from intermediate theorem 4.26 that if r ∈ [s, t] then there existsc ∈ I such that

r = f(c) ∈ f(I).

Therefore [s, t] ⊆ f(I), by theorem 4.27, f(I) is an interval.

Corollary 4.31 Let f : [a, b] −→ R be continuous on [a, b]. For d ∈ R, if

inf {f(x) : x ∈ [a, b]} ≤ d ≤ sup {f(x) : x ∈ [a, b]}

then there exists a number c ∈ [a, b] such that

f(c) = d.

To end this section, we introduce an important fixed point theorem (see Fig. 4.4). Anumber of important results can be proved on the basis of existence of fixed points offunctions so it is of importance to have some affirmative criteria.

4.3 - Uniformly Continuous Functions 73

Definition 4.32 (Fixed point) Let f : D −→ R be a continuous function. A point xis said to be a fixed point of f in case

f(x) = x.

Theorem 4.33 (Fixed point theorem) Let f : [0, 1] −→ [0, 1] be a continuous func-tion then there exists a point c ∈ [1, 0] such that

f(c) = c.

Proof. If f(0) = 0 and f(1) = 1 then there exists a fixed point. So, let us assume thatf(0) 6= 0 and f(1) 6= 1. For x ∈ [0, 1], let us define

g(x) := f(x)− x

then1. g is continuous2. g(0) = f(0) > 0

3. g(1) = f(1)− 1 < 0.Therefore, by intermediate value theorem 4.26, there exists c ∈ [0, 1] such that

g(c) = 0 so that f(c) = c.

0 1

1

c

c

y=x

y=f (x)

Fig. 4.2 – Description of fixed point theorem

4.3 Uniformly Continuous Functions

In this section, we introduce very important notion of uniform continuity. The dis-tinction between continuity and uniform continuity is somewhat subtle and was notfully appreciated until the work of Weierstrass and the mathematicians of his era, butit proved be very significant in applications.

Let f : D −→ R be a continuous function. Definition 4.1 states that the followingstatements are equivalent :


1. f is continuous at every point a ∈ D.2. Given ε > 0 and a ∈ D, there exists δ(ε, a) > 0 such that for all x ∈ D and|x− a| < δ(ε, a) then

|f(x)− f(a)| < ε.

The point we wish to emphasize here is that δ depends, in general, on both ε > 0 anda ∈ A. The fact that δ depends on a is a reflection of the fact that the function f maychange its values rapidly near certain points and slowly near other points (for example,f(x) := sin(x−1) for x > 0, refer to Fig. 3.4).

Now it often happens that the function f is such that the number δ can be chosento be independent of the point a ∈ D and to depend only on ε > 0 :

Example 4.34 Let f(x) := 2x for x ∈ R. Then for ε > 0, we can choose δ(ε, a) := ε2

such that if |x− a| < δ then

|f(x)− f(a)| = 2|x− a| < ε.

Therefore, f is continuous on R. Notice that, in this example, δ depends only on ε > 0.

On the other hand, consider the following example :

Example 4.35 (Example 3.11 revisited) Let f(x) := x2 for all x ∈ R. We want tomake the difference as

|f(x)− a2| = |x2 − a2| < ε

for a preassigned ε > 0. To do so, we note that x2 − a2 = (x + a)(x − a). Moreover if|x− a| < 1 then

|x| ≤ |a|+ 1 so that |x + a| ≤ |x|+ |a| ≤ 2|a|+ 1.

Therefore, if |x− a| < 1, we have

|f(x)− a2| = |x2 − a2| = |x + a||x− a| ≤ (2|a|+ 1)|x− a|. (4.4)

The last term of above inequality will be less than ε provided we take

|x− a| < ε

2|a|+ 1.


δ(ε, a) := min

{1,

ε

2|a|+ 1

},

then if 0 < |x−a| < δ, it follow first that |x−a| < 1 so that (4.4) is valid, and therefore,

|f(x)− a2| = |x2 − a2| = |x + a||x− a| ≤ (2|a|+ 1)ε

2|a|+ 1< ε.

Therefore, f is continuous on R. Notice that, in this example, δ depends on both ε > 0and a ∈ A. For example, let us take ε = 1

4. If a = 3 and a = 10 then δ = 1

28and δ = 1

84,

respectively.


Definition 4.36 (Uniformly continuous function) We say that f : D −→ R isuniformly continuous on D if for each ε > 0 there exists a δ(ε) > 0 such that if|x− y| < δ, x, y ∈ D then

|f(x)− f(y)| < ε.

It is clear that if f is uniformly continuous on D, then it is continuous at everypoint of D. In general, however, the converse does not hold, e.g., f(x) = x−1 on the setD := {x ∈ R : x > 0}.

It is useful to formulate a condition equivalent to saying that f is not uniformlycontinuous on D. We give such criteria in the next result.

Theorem 4.37 (Nonuniform continuity criteria) Let f : D −→ R be a functionthen the following statements are equivalent :

1. f is not uniformly continuous on D.2. There exists an ε0 > 0 such that for every delta δ > 0 there are points x, y ∈ D

such that|x− y| < δ and |f(x)− f(y)| ≥ ε0.

3. There exists an ε0 > 0 and two sequences {xn} and {yn} in D such that

limn→∞

(xn − yn) = 0 and |f(xn)− f(yn)| ≥ ε0

for all n ∈ N.

Example 4.38 Prove that

f(x) =1

x

is not uniformly continuous on (0, 1).

Proof. Let 0 < ε0 ≤ 1. For every 0 < δ < 1, let us choose x, y ∈ (0, 1) such that

x =δ

10and y =

δ

11.

Then clearly x, y ∈ (0, 1),

|x− y =

∣∣∣∣ δ

10− δ

11

∣∣∣∣ =δ

110< δ

and|f(x)− f(y)| =

∣∣∣∣1x − 1

y

∣∣∣∣ =

∣∣∣∣10

δ− 11

δ

∣∣∣∣ =1

δ> 1 ≥ ε0.

Therefore, by theorem 4.37, f is not uniformly continuous on (0, 1).

We now present an important result that assures that a continuous function on aclosed bounded interval I is uniformly continuous on I.

Theorem 4.39 (Uniform continuity theorem) Let I = [a, b] be a closed intervaland f : I −→ R be a continuous function on I. Then f is uniformly continuous on I.


Proof. Assume that f is not uniformly continuous on I then, by theorem 4.37, thereexists ε0 > 0 and two sequences {xn} and {yn} in I such that

|xn − yn| <1

nand |f(xn)− f(yn)| ≥ ε0

for all n ∈ N. Since I is bounded interval and {xn} is bounded sequence ; by the Bolzano-Weierstrass theorem, there exists a subsequence {xnk

} of {xn} such that

limk→∞

xnk= c.

Since I is closed, by theorem 2.14, c ∈ I. It is clear that the corresponding subsequence{ynk

} of {yn} also satisfieslimk→∞

ynk= c

since|ynk

− c| ≤ |ynk− xnk

|+ |xnk− c|.

Now if f is continuous at c ∈ I, then both of the sequences {f(xnk)} and {f(ynk

)}satisfies

limk→∞

f(xnk) = c and lim

k→∞f(ynk

) = c.

But this is impossible since|f(xn)− f(yn)| ≥ ε0

for all n ∈ N. Therefore, Then f is uniformly continuous on I.

If a uniformly continuous function is given on a set that is not closed boundedinterval, then it is sometimes difficult to establish its uniform continuity. However, thereis a condition that frequently occurs that is sufficient to guarantee uniform continuity.Now, we will consider this condition.

Definition 4.40 (Lipschitz function) Let f : D −→ R be a function. If there existsa constant K > 0 such that

|f(x)− f(y)| ≤ K|x− y| (4.5)

for all x, y ∈ D, then f is said to be a Lipschitz function or to satisfy a Lipschitzcondition on D.

Notice that the condition (4.5) that a function f : D −→ R on an interval I is aLipschitz function can be interpreted geometrically as follows : If we write the conditionas ∣∣∣∣f(x)− f(y)

x− y

∣∣∣∣ ≤ K

for x, y ∈ I, x 6= y, then the quantity inside the above absolute values is the slope ofa line segment joining the points (x, f(x)) and (y, f(y)). Thus a function f satisfies aLipschitz condition if and only if the slopes of all line segments joining two points onthe graph of f over I are bounded by some number K.

Theorem 4.41 If f : D −→ R is a Lipschitz function, then f is uniformly continuouson D.


Proof. Since f is a Lipschitz function, there exists a constant K > 0 such that

|f(x)− f(y)| ≤ K|x− y|

for all x, y ∈ D. Then given ε > 0, we can take

δ :=ε

K.

If x, y ∈ D satisfy |x− y| < δ, then

|f(x)− f(y)| ≤ K|x− y| < Kδ = Kε

K= ε.

Therefore, f is uniformly continuous on D.

Example 4.42 If f(x) := x2 on I = [0, 2] then f is uniformly continuous on I, referto Fig. 4.3.

Proof. For x, y ∈ I, there exists K = 4 > 0 such that

|f(x)− f(y)| = |x2 − y2| = |x + y||x− y| ≤ (|x|+ |y|)|x− y| ≤ 4|x− y|.

Therefore, f is a Lipschitz function. By theorem 4.41, f is uniformly continuous on I.

Remark 4.43 There are some remarks :1. Let us notice that f(x) := x2 is uniformly continuous on a closed bounded interval,

refer to example 4.42, but f does not satisfy a Lipschitz condition on the interval[0,∞).

2. Not every uniformly continuous function is a Lipschitz function. For example,let g(x) :=

√x for x ∈ [0, 1]. Since g is continuous on [0, 1], it follows from the

uniform continuity theorem (theorem 4.39) that g is uniformly continuous functionon [0, 1]. However, there is no K > 0 such that

|g(x)| ≤ K|x|

for all x ∈ [0, 1]. Therefore, g is not a Lipschitz function on [0, 1], refer to Fig.4.3.

3. The uniform continuity theorem (theorem 4.39) and theorem 4.41 can sometimesbe combined to establish the uniform continuity of a function on a set.

We have seen examples of functions that are continuous but not uniformly conti-nuous on an open intervals, e.g., f(x) = x−1 on (0, 1). On the other hand, by theuniform continuity theorem, a function that is continuous on a closed bounded intervalis always uniformly continuous. So the question arises : under what conditions is a func-tion uniformly continuous on a bounded open interval ? The answer reveals the strengthof uniform continuity, for it will be shown that a function on an open interval (a, b)is uniformly continuous if and only if it can be defined at the endpoints to produce afunction that is continuous on the closed interval. We first establish a result that is ofinterest in itself.


−0.5 0 0.5 1 1.5 2 2.50

1

2

3

4

5

6

7

x−axis

y−ax

is

(a) Graph of f(x) = x2

0 0.5 1 1.50

0.2

0.4

0.6

0.8

1

1.2

1.4

x−axis

y−ax

is

(b) Graph of g(x) =√

x

Fig. 4.3 – Example of Lipschitz and non Lipschitz functions

Theorem 4.44 If If f : D −→ R is uniformly continuous on D and if {xn} is a Cauchysequence in D, then {f(xn)} is a Cauchy sequence in R.

Proof. Since f is uniformly continuous on D, for given ε > 0, there exists δ > 0 suchthat if |x− y| < δ, x, y ∈ D then

|f(x)− f(y)| < ε.

Since {xn} is a Cauchy sequence in D, there exists N ∈ N such that for all n, m ≥ N ,

|xm − xn| < δ.

By the choice of δ, this implies that for n, m ≥ N , we have

|f(xm)− f(xn)| < ε.

Therefore, {f(xn)} is a Cauchy sequence in R.

The preceding result gives us an alternative way of seeing that f(x) := x−1 is notuniformly continuous on (0, 1). Note that the sequence given by xn = n−1 in (0, 1) is aCauchy sequence but the image sequence given by f(xn) = n is not a Cauchy sequence.

Theorem 4.45 (Continuous extension theorem) A function f : (a, b) −→ R isuniformly continuous on (a, b) if and only if it can be defined at the endpoints a and bsuch that the extended function f ∗ is continuous on [a, b].

Proof. If a function f can be defined at the endpoints a and b such that the extendedfunction f ∗ is continuous on [a, b], by the uniform continuity theorem (theorem 4.39), fis uniformly continuous.


Conversely, suppose that f is uniformly continuous on (a, b). We shall show how toextend f to a. If {xn} is a sequence in (a, b) with

limn→∞

xn = a

then it is a Cauchy sequence by lemma 2.36, so that, by preceding theorem, {f(xn)} isalso a Cauchy sequence, and so is convergent by Cauchy convergent criterion (theorem2.38). Thus the limit

limn→∞

f(xn) = p

exists. If {yn} is any other sequence in (a, b) that converges to a then

limn→∞

(yn − xn) = a− a = 0

so by the uniform continuity of f , we have

limn→∞

f(yn) = limn→∞

{f(yn)− f(xn)}+ limn→∞

f(xn) = 0 + p = p.

Since we get the same value p for every sequence converging to a, we infer that

limx→a

f(x) = p.

Therefore, if we define f(a) = p then f is continuous at a. The same argument appliesto b as follows :

limx→b

f(x) = q.

So, we conclude that f has a continuous extension f ∗ to the closed interval [a, b] suchthat

f ∗(x) =

f(x) if x ∈ (a, b)

p if x = aq if x = b.

We now consider the one-sided continuity.

Definition 4.46 (One-sided continuous function) Let f : D −→ R be a functionand a ∈ D.

1. We say that f is right-continuous function at a if for every ε > 0 there existsδ > 0 such that for all x ∈ D with a < x < a + δ then

|f(x)− f(a)| < ε.

2. We say that f is left-continuous function at a if for every ε > 0 there existsδ > 0 such that for all x ∈ D with a− δ < x < a then

|f(x)− f(a)| < ε.

Theorem 4.47 Let f : (a, b) −→ R be a function and c ∈ (a, b). Then f is right-continuous function at c if and only if there exists f(c+) and f(c+) = f(c).

Theorem 4.48 Let f : (a, b) −→ R be a function and c ∈ (a, b). Then f is left-continuous function at c if and only if there exists f(c−) and f(c−) = f(c).


Theorem 4.49 Let f : (a, b) −→ R be a function and c ∈ (a, b). Then the followingstatements are equivalent :

1. f is continuous at c.2. There exists f(c+) and f(c−). Moreover, f(c+) = f(c) = f(c−).

Proof. Suppose that f is continuous at c then clearly there exists f(c+) and f(c−)since

limx→c

f(x) = f(c)

so that f(c+) = f(c) = f(c−).

Conversely, suppose that f(c+) = f(c) = f(c−) for c ∈ (a, b). Then by the definition4.46, for given ε > 0 there exists δ1 > 0 such that if c < x < c + δ1 then

|f(x)− f(c)| < ε.

Moreover, there exists δ2 > 0 such that if c− δ2 < x < c then

|f(x)− f(c)| < ε.

Now, let δ = min {δ1, δ2} then if |x− c| < δ then

|f(x)− f(c)| < ε.

Therefore, f is continuous at c.

Definition 4.50 Let f : D −→ R be a function and c ∈ D.1. We say that f is discontinuous at a if f does not defined at a or there exists

limx→a

f(x) but does not equal to f(a). In this case, the point a is called removablediscontinuous point.

2. We say that f is jump discontinuous at a if there exists f(a+) and f(a−) butf(a+) 6= f(a−).

Example 4.51 Let g : R −→ R be a function defined as (refer to Fig. 4.4)

g(x) :=

x2 − 4

x− 2if x 6= 2

2 if x = 2

Then g is discontinuous at x = 2 because

limx→2

g(x) = 4 6= 2.

Therefore, x = 2 is removable discontinuous point.

Example 4.52 Let f : [0,∞) −→ R be a function defined as (refer to Fig. 4.4)

g(x) :=

{3− x2 if x > 1

x if 0 ≤ x ≤ 1


0 1 2 3 41

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

x−axis

y−ax

is

(a) Graph of g(x) (example 4.51)

0 0.5 1 1.5 2−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

x−axis

y−ax

is

(b) Graph of f(x) (example 4.52)

Fig. 4.4 – Example of discontinuous and jump discontinuous functions

Since f(1−) = 1, f(1+) = 2 and f(1+) 6= f(1−), f is jump discontinuous at x = 1.

Definition 4.53 (Increasing, decreasing and monotone function) Letf : D −→ R be a function.

1. f is said to be increasing on D if whenever x1, x2 ∈ D and x1 ≤ x2 then f(x1) ≤f(x2).

2. f is said to be decreasing on D if whenever x1, x2 ∈ D and x1 ≤ x2 then f(x1) ≥f(x2).

3. f is said to be strictly increasing on D if whenever x1, x2 ∈ D and x1 ≤ x2 thenf(x1) < f(x2).

4. f is said to be strictly decreasing on D if whenever x1, x2 ∈ D and x1 ≤ x2 thenf(x1) > f(x2).

5. If f is either increasing or decreasing on D, we say that f is monotone on D.6. If f is either strictly increasing or strictly decreasing on D, we say that f is

strictly monotone on D.

Theorem 4.54 Let I ⊆ R be an open interval and let f : I −→ R be an increasingfunction on I. Then for all c ∈ I there exists f(c+), f(c−) and

sup {f(x) : x < c, x ∈ I} = f(c−) ≤ f(c) ≤ f(c+) = inf {f(x) : c < x, x ∈ I} .

Moreover, if c, d ∈ I satisfies c < d then

f(c+) ≤ f(d−).

Proof. Note that if x ∈ I and x < c then f(x) ≤ f(c). Hence the set

{f(x) : x < c, x ∈ I}


is nonempty and bounded above by f(c). Thus the indicated supremum

sup {f(x) : x < c, x ∈ I}

exists ; we denote it by L. If ε > 0 is given then L− ε is not an upper bound of this set.Hence there exists xc ∈ I, xc < c such that

L− ε < f(xc) ≤ L.

Since f is an increasing function, we deduce that if δ := c−xc and if 0 < c−x < δ thenxc < x < c so that

L− ε < f(xc) ≤ f(x) ≤ L.

Therefore |f(x)− L| < ε when 0 < c− x < δ ; so that

limx→c−

f(x) = f(c−) = L implies sup {f(x) : x < c, x ∈ I} = f(c−) ≤ f(c).

Similarly,f(c) ≤ f(c+) = inf {f(x) : c < x, x ∈ I} .

Let c, d ∈ I and c < d then

f(c+) = inf {f(x) : c < x, x ∈ I}≤ inf {f(x) : c < x < d}≤ sup {f(x) : c < x < d}≤ sup {f(x) : x < d, x ∈ I} = f(d−).

The next result gives criteria for the continuity of an increasing function f at a pointc that is not an endpoint of the interval on which f is defined.

Corollary 4.55 Let I ⊆ R be an interval and let f : I −→ R be an increasing functionon I. Suppose that c ∈ I is not an endpoint of I then the following statements areequivalent.

1. f is continuous at c.2. f(c−) = f(c) = f(c+).3. sup {f(x) : x < c, x ∈ I} = f(c−) = f(c) = f(c+) = inf {f(x) : c < x, x ∈ I} .

We now consider the existence of inverses for functions that are continuous on aninterval I ⊆ R. We recall that a function f : I −→ R has an inverse function if andonly if f is injective (one-one) ; that is, x, y ∈ I and x 6= y imply that f(x) 6= f(y).Note that a strictly monotone function is injective and so has an inverse. In the nexttheorem, we show that if f : I −→ R is a strictly monotone continuous function then fhas an inverse function f−1 on J := f(I) that is strictly monotone and continuous onJ . In particular, if f is strictly increasing then so is f−1, and if f is strictly decreasingthen so is f−1.

Theorem 4.56 (Continuous inverse theorem) Let I ⊆ R be an interval and f :I −→ R be strictly increasing (or decreasing) and continuous on I. Then the functionf−1 inverse to f strictly increasing (or decreasing) and continuous on J := f(I).


Proof. Since f is continuous and I is an interval, it follows from the preservation ofintervals theorem (theorem 4.30) that J := f(I) is an interval. Moreover, since f isstrictly increasing on I, it is injective on I. Therefore the function f−1 : J −→ R inverseto f exists.

In order to show that the inverse function f−1 is strictly increasing, let y1, y2 ∈ Jwith y1 < y2. Then there exists x1, x2 ∈ I such that

y1 = f(x1) and y2 = f(x2).

Let us assume that x1 ≥ x2, which implies that

y1 = f(x1) ≤ f(x2) = y2,

contrary to the hypothesis that y1 < y2. Therefore we have

f−1(y1) = x1 < x2 = f−1(y2),

i.e., f−1 is strictly increasing.

It remains to show that f−1 is continuous on J . Let us assume that f−1 is discon-tinuous at a point a ∈ J . Then the jump of f−1 at a is nonzero so that by theorem4.54,

limy→a−

f−1(y) = f−1(a−) < f−1(a+) = limy→a+

f−1(y).

If we choose any number x 6= f−1(a) satisfying

limy→a−

f−1(y) < x < limy→a+

f−1(y),

then for any y ∈ J ,x 6= f−1(y) (refer to Fig. 4.5).

Hence x 6= I, which contracts the fact that I is an interval. Therefore, f−1 is continuouson J .

Ja

x f (a)-1

Fig. 4.5 – f−1(y) 6= x for y ∈ J .

In advanced analysis and topology, the notion of a compact set is of enormous im-portance. This is less true in R because the Heine-Borel theorem gives a very simplecharacterization of compact sets in R. Nevertheless, the definition and the techniquesused in connection with compactness are very important, and the real line provides anappropriate place to see the idea of compactness for the first time.


Definition 4.57 (Open cover) Let I be an indexed set and Ii be an open interval fori ∈ I. An open cover of [a, b] is a collection I := {Ii : i ∈ I} of open sets in R whoseunion contains [a, b] ; that is,

[a, b] ⊆⋃i∈I

Ii.

If J is a subcollection of sets from I such that the union of the sets in J also contains[a, b] then J is called a subcover of I. If J consists of finitely many sets then we callJ a finite subcover of I.

Example 4.58 Let A := [1,∞) then the following collections of sets are all open coversof A, refer to [3].

1. I1 := {(0,∞)}.2. I2 := {(n− 1, n + 1) : n ∈ N}.3. I3 := {(0, n) : n ∈ N}.

Theorem 4.59 (Heine-Borel theorem) Let I := {Ii : i ∈ I} be a collection of openintervals Ii such that

[a, b] ⊆⋃i∈I

Ii.

Then there exists a finite subset J :={Iαj

: j = 1, 2, · · · , n}

of I such that

[a, b] ⊆n⋃

j=1

Iαj= Iα1 ∪ Iα2 ∪ · · · ∪ Iαn .

Proof. See the theorem 4.29 of main textbook.


1. Prove that if f : (a, b) −→ R is continuous at c ∈ (a, b) and f(c) > 0 then thereexists δ > 0 such that if x ∈ (a, b) satisfies |x− c| < δ then

f(x) > 0.

2. Give an example of a function f : [0, 1] −→ R that is discontinuous at every pointof [0, 1] but such that |f | is continuous on [0, 1].

3. Suppose that f : [0, 1] −→ R is continuous on [0, 1] and that f(0) < g(0), f(1) >g(1). Prove that there exists a point x ∈ (0, 1) such that

f(x) = g(x).

4. Show that the equation x = cos x has at least one root in the interval(0, π

2

).

(a) By theorem 4.26 (theorem 4.9 of main textbook).

(b) By theorem 4.32 (theorem 4.13 of main textbook).

RÉFÉRENCES 85

5. A function f : R −→ R is said to be periodic on R if there exists a number a > 0such that

f(x + a) = f(x)

for all x ∈ R. Prove that a continuous periodic function on R is bounded anduniformly continuous on R.

6. If a function f is defined on R by

f(x) :=

[x]

xfor x > 0

x[x] for x ≤ 0.

Prove that f is continuous at x = 0.7. Discriminate the Lipschitz function :

(a) f(x) := x2, x ∈ (−2, 1].

(b) g(x) := x sin1

x, x ∈ (0, 1].

(c)√

x, x ∈ [a,∞) for a > 0.

Références

[1] R. G. Bartle, The Elements of Integration and Lebesgue measure, John Wiley &Sons, Inc., 1966.

[2] R. G. Bartle and D. R. Sherbert, The Elements of Real Analysis, Second Edition,John Wiley & Sons, Inc., 1976.

[3] R. G. Bartle and D. R. Sherbert, Introduction to Real Analysis, Third Edition,John Wiley & Sons, Inc., 2000.

[4] R. Johnsonbaugh and W. E. Pfaffenberger, Foundations of Mathematical Analysis,Dekker, 1981.

[5] J. E. Marsden and M. J. Hoffman, Elementary Classical Analysis, Second Edition,Freeman, 1993.

[6] C. W. Patty, Foundations of Topology, PWS Publishing Company, 1993.[7] W. Rudin, Principles of Mathematical Analysis, Third Edition, McGraw–Hill,

1976.

Documents

Introduction to Mathematical Analysis