Chapter 1: Limits and Continuityfelix_kwok/2018/chapter1_notes1.pdf · Chapter 1: Limits and Continuity Spring 2018 Department of Mathematics Hong Kong Baptist University 1/75. x1.1

Chapter 1: Limits and Continuity

Spring 2018

Department of MathematicsHong Kong Baptist University

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§1.1 Examples where limits arise

Calculus has two basic procedures: differentiation andintegration. Both procedures are based on the fundamentalconcept of the limit of a function.

It is the idea of limit that distinguishes Calculus from Algebra,Geometry, and Trigonometry, which are useful for describingstatic situations.

This section considers some examples of phenomena wherelimits arise in a natural way.

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Example 1 :

Consider the area of a circle of radius r . In high school we aretaught that the area A of the circle is

A = πr2.

The deduction of this area formula lies in regarding the circle as a“limit” of regular polygons. Suppose a regular polygon having nsides is inscribed in the circle of radius r , and let An be the area ofthe polygon. It is clear that An is less than A for each n. But if nis large, An is close to A. That is, we would expect that An

approaches the limit A when n goes to infinitely large.

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The area An of the polygon has the following expression:

An = r2n · sin(π

n) cos(

π

n).

Then by the trigonometric formula,

An =r2n

2sin(

2π

n).

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n t = 2π/n sin(t)/t4 1.5707963 0.636628 0.7853982 0.900316

16 0.3926991 0.97449532 0.1963495 0.99358764 0.0981748 0.998394

128 0.0490874 0.999598256 0.0245437 0.9999512 0.0122718 0.999975

1024 0.0061359 0.9999942048 0.003068 0.999998

Finally, noting that sin(t)/t → 1 as t → 0, we have

An = πr21

2π/nsin(2π/n)→ πr2 as n→∞.

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§1.2 Limits of Functions

The concept of limit is the cornerstone on which thedevelopment of Calculus rests.

Intuitively speaking, the limit process involves examining thebehavior of a function f (x) as x approaches a number c thatmay or may not be in the domain of f .

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Example 2 :

Describe the behavior of the function f (x) =x2 − 4

x − 2near x = 2.

x f(x)1.9 3.91.99 3.991.999 3.9991.9999 3.99991.99999 3.999992.00001 4.000012.0001 4.00012.001 4.0012.01 4.012.1 4.1

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Example 2 :

Describe the behavior of the function f (x) =x2 − 4

x − 2near x = 2.

Solution: Note that f (x) is defined for all real numbers x exceptx = 2. For any x 6= 2 we have

f (x) =(x + 2)(x − 2)

x − 2= x + 2 for x 6= 2.

The graph of f is the line y = x + 2 with the point (x , y) = (2, 4)removed. Though f (2) is not defined, it is clear that f (x) becomescloser to 4 as x approaches 2. In other words, we say that f (x)approaches the limit 4 as x approaches 2. We write this as

limx→2

f (x) = limx→2

x2 − 4

x − 2= 4.

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(Informal) definition of the limit

Definition

If f (x) is defined for all x near c , except possibly at c itself, and ifwe can ensure that f (x) is as close as we want to L by taking xclose enough to (but different from) c, we say that the function fapproaches the limit L as x approaches c , and we write it as

limx→c

f (x) = L.

Warning: limx→c f (x) does not dependon whether f (c) exists, or what value itis. It only depends on values near x = c .For the function g(x), we have

limx→2

g(x) = 2, but g(2) = 1.

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Limits of two linear functions

(1) For the identity function f (x) = x , we have

limx→c

f (x) = limx→c

x = c .

That is, the limit of f (x) = x is c as x approaches c .

(2) For any constant k, we have

limx→c

k = k .

That is, the limit of a constant is the constant itself.

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Limits of two linear functions

(1) For the identity function f (x) = x , we have

limx→c

f (x) = limx→c

x = c .

That is, the limit of f (x) = x is c as x approaches c .

(2) For any constant k, we have

limx→c

k = k .

That is, the limit of a constant is the constant itself.

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Limit Rules

If limx→c

f (x) = L, limx→c

g(x) = M, and k is a constant, then

1) limx→c

kf (x) = kL.

2) limx→c

[f (x)± g(x)] = L±M.

3) limx→c

[f (x) · g(x)] = LM.

4) limx→c

f (x)

g(x)=

L

Mif M 6= 0.

5) If f (x) ≤ g(x) on an interval containing c in its interior, thenthe order of the limits is preserved, i.e., L ≤ M.

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Example 3 :

Find the following limits:

(a) limx→2

x2 − 1

x − 1, (b) lim

x→1

x2 − 1

x − 1.

Solution: (a) Considering the numerator and denominatorseparately, we have

limx→2

(x2 − 1)(2)= lim

x→2x2 − lim

x→21

(3)=(

limx→2

x)(

limx→2

x)− 1 = 3

and

limx→2

(x − 1)(2)=(

limx→2

x)− 1 = 2− 1 = 1.

Since the denominator does not tend to zero, we can use rule(4) to conclude that

limx→2

x2 − 1

x − 1

(4)=

limx→2(x2 − 1)

limx→2(x − 1)=

3

1= 3.

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Example 3 :


(a) limx→2

x2 − 1

x − 1, (b) lim

x→1

x2 − 1

x − 1.

Solution: (b) Here, we cannot use rule (4) directly because thedenominator tends to zero, i.e.,

limx→1

(x − 1) = 0.

But when x 6= 1, we have

x2 − 1

x − 1=

(x − 1)(x + 1)

x − 1= x + 1,

so

limx→1

x2 − 1

x − 1= lim

x→1(x + 1)

(2)= 1 + 1 = 2.

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Remark 1: For ease of calculation, one typically writes for part (a)

limx→2

x2 − 1

x − 1

(4)=

limx→2(x2 − 1)

limx→2(x − 1)

(2,3)=

3

1= 3.

But the first step is only valid because limx→2(x − 1) = 1 6= 0. Ifwe tried to apply the same calculation to (b), we would get

limx→1

x2 − 1

x − 1

(4)=

limx→1(x2 − 1)

limx→1(x − 1)

(2,3)=

0

0=??

When this happens, the calculation is invalid, and we need tosimplify before substituting numbers into the expressions.

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Remark 2: For any positive integer n, by property 3) we have

limx→c

[f (x)]n = Ln.

More generally, for any non-integer t such that Lt exists, we stillhave

limx→c

[f (x)]t = Lt .

Remark 3: Let f (x) = x , we have limx→c

xn = cn. Let P(x) and Q(x)

be any two polynomials, then

(i) limx→c

P(x) = P(c),

(ii) limx→c

P(x)

Q(x)=

P(c)

Q(c)if Q(c) 6= 0.

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Example 4 :

Find limx→1

x3 − 4x + 3

x2 + 5x − 6.

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Example 4 :

Find limx→1

x3 − 4x + 3

x2 + 5x − 6.

Wrong answer:

limx→1

x3 − 4x + 3

x2 + 5x − 6=

limx→1(x3 − 4x + 3)

limx→1(x2 + 5x − 6)=

0

0=?

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Example 4 :

Find limx→1

x3 − 4x + 3

x2 + 5x − 6.

Correct answer: Both numerator and denominator are divisible by(x − 1). Long division then gives

limx→1

x3 − 4x + 3

x2 + 5x − 6= lim

x→1

(x − 1)(x2 + x − 3)

(x − 1)(x + 6)

= limx→1

x2 + x − 3

x + 6

=limx→1(x2 + x − 3)

limx→1(x + 6)= −1

7.

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Example 5 :

Find limx→2

x −√

2 + x

x − 2.

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Example 5 :

Find limx→2

x −√

2 + x

x − 2.

Wrong answer:

limx→2

x −√

2 + x

x − 2=

limx→2(x −√

2 + x)

limx→2(x − 2)=

0

0=?

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Example 5 :

Find limx→2

x −√

2 + x

x − 2.

Correct answer:

limx→2

x −√

2 + x

x − 2= lim

x→2

(x −√

2 + x)(x +√

2 + x)

(x − 2)(x +√

2 + x)

= limx→2

x2 − x − 2

(x − 2)(x +√

2 + x)

= limx→2

(x − 2)(x + 1)

(x − 2)(x +√

2 + x)

= limx→2

x + 1

x +√

2 + x=

3

4.

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Left and Right Limits

Definition

(a) If f (x) is defined on some interval (a, c) extending to the leftof x = c , and if we can ensure that f (x) is as close as wewant to L by taking x to the left of c and close enough to c,then we say f (x) has left limit L at x = c , and we write it as

limx→c−

f (x) = L.

(b) If f (x) is defined on some interval (c , b) extending to the rightof x = c , and if we can ensure that f (x) is as close as we wantto L by taking x to the right of c and close enough to c , thenwe say f (x) has right limit L at x = c , and we write it as

limx→c+

f (x) = L.

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Example 6 :

(a) Find the left and right limits of f (x) =x2 − 4

x − 2at x = 2;

(b) Find the left and right limits of g(x) = x/|x | at x = 0.

Solution:

(a) For f (x), we have

limx→2−

f (x) = 4 and limx→2+

f (x) = 4.

(b) For g(x), we have

limx→0−

g(x) = −1 and limx→0+

g(x) = 1.

Note that this is the so-called sign function, i.e.,sgn(x) = x/|x |.

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Example 6 :


x − 2at x = 2;


Solution:


limx→2−


f (x) = 4.


limx→0−


g(x) = 1.


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Example 6 :


x − 2at x = 2;


Solution:


limx→2−


f (x) = 4.


limx→0−


g(x) = 1.


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Relationship between one-sided and two-sided limits

Theorem

[Existence of a Limit] The two-sided limit limx→c

f (x) exists if and

only if both one-sided limits limx→c−

f (x) and limx→c+

f (x) exist and are

equal to each other. In this case, we have

limx→c

f (x) = limx→c−

f (x) = limx→c+

f (x).

Note: If the two one-sided limits do not agree, or if one of themdoes not exist, then the two-sided limit does not exist. Forexample,

limx→0

x

|x |does not exist.

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The Squeeze Theorem

Theorem

Suppose that f (x) ≤ g(x) ≤ h(x) holds for all x in some openinterval containing a, except possibly at x = a itself. Suppose alsothat

limx→a

f (x) = limx→a

h(x) = L.

Then limx→a g(x) = L too. Similar statements hold for left andright limits.

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The Squeeze Theorem

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Example 7 :

(a) Given that 3− x2 ≤ u(x) ≤ 3 + x2 for all x 6= 0, find limx→0

u(x).

(b) If limx→a|f (x)| = 0, find lim

x→af (x).

Solution:

(a) Note that limx→0

(3− x2) = 3 and limx→0

(3 + x2) = 3. By the

Squeeze theorem, we have limx→0

u(x) = 3.

(b) Note that −|f (x)| ≤ f (x) ≤ |f (x)| and

limx→a{−|f (x)|} = − lim

x→a|f (x)| = 0.

By the Squeeze theorem, we have limx→a

f (x) = 0.

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Example 7 :


u(x).


x→af (x).

Solution:


(3− x2) = 3 and limx→0

(3 + x2) = 3. By the


u(x) = 3.



x→a|f (x)| = 0.


f (x) = 0.

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Example 7 :


u(x).


x→af (x).

Solution:


(3− x2) = 3 and limx→0

(3 + x2) = 3. By the


u(x) = 3.



x→a|f (x)| = 0.


f (x) = 0.

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Example 8 :

Calculate limx→0

x sin

(1

x

).

Solution:

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Example 8 :

Calculate limx→0

x sin

(1

x

).

Solution:

Note that | sin(1/x)| ≤ 1 for all values of x . Thus, we have

−|x | ≤ x sin

(1

x

)≤ |x |.

Since limx→0 |x | = limx→0−|x | = 0, by the Squeeze theorem, weget

limx→0

x sin

(1

x

)= 0.

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f (x) = x sin

(1

x

)

−1 −0.8 −0.6 −0.4 −0.2 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

x

x*si

n(1/

x)

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Note: limx→0

sin

(1

x

)does NOT exist!

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

x

sin(

1/x)

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Example 9 (See also §2.5):

Calculate limθ→0

sin θ

θ(where θ is in radians).

Solution:

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Calculate limθ→0

sin θ


Solution:

For 0 < θ < π/2,

Area of sector =θ

2

Area of 4OPB =1

2sin θ cos θ

Area of 4OTA =1

2tan θ

Hence,

sin θ cos θ

2≤ θ

2≤ tan θ

2.

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Calculate limθ→0

sin θ


Solution:

(i) Thus, for 0 < θ < π/2, we have

cos θ ≤ sin θ

θ≤ 1

cos θ

with

limθ→0+

cos θ = 1, limθ→0+

1

cos θ= 1.

By the Squeeze theorem, we obtain limθ→0+

sin θ

θ= 1.

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Calculate limθ→0

sin θ


Solution:

(ii) For θ < 0, we can substitute α = −θ to get

limθ→0−

sin θ

θ= lim

α→0+

sin(−α)

(−α)= lim

α→0+

− sin(α)

−α= 1.

(iii) Thus, combining the two one-sided limits, we get

limθ→0

sin θ

θ= 1.

Note: This is an important result and should be memorized.

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Example 10 :

Calculate limx→0

1− cos(x)

x2.

Solution:

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Example 10 :

Calculate limx→0

1− cos(x)

x2.

Solution:

Using the trigonometric identity cos(x) = 1− 2 sin2(x/2), wededuce that

limx→0

1− cos(x)

x2= lim

x→0

2 sin2(x/2)

x2

= limx→0

2 sin2(x/2)

4 · (x/2)2

=1

2

(limx→0

sin(x/2)

(x/2)

)·(

limx→0

sin(x/2)

(x/2)

)=

1

2.

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§1.3 Limits at Infinity and Infinite Limits

In this section, we consider several scenarios not covered by thelimit definitions in Section 1.2:

(i) limit at infinity: when x becomes arbitrarily large positive(denoted by x →∞).

Remark: The symbol ∞, called “infinity”, does not representa real number. We cannot use ∞ in arithmetic in the usualway. For instance, we cannot say ∞+ 1 >∞.

(ii) limit at negative infinity: when x becomes arbitrarily largenegative (denoted by x → −∞).

(iii) infinite limits: they are not really limits at all but provideuseful symbolism for describing the behavior of functionswhose values become arbitrarily (positive or negative) large.

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Definition

(a) If f (x) is defined on an interval (a,∞) and if we can ensurethat f (x) is as close as we want to the number L by taking xlarge enough, then we say that f (x) approaches the limit Las x approaches infinity, and we write

limx→∞

f (x) = L.

(b) If f (x) is defined on an interval (−∞, b) and if we can ensurethat f (x) is as close as we want to the number M by taking xnegative and large enough in absolute value, then we say thatf (x) approaches the limit M as x approaches negativeinfinity, and we write

limx→−∞

f (x) = M.

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Example 11 :

(a) Evaluate limx→∞

f (x) and limx→−∞

f (x) for f (x) =x√

x2 + 1.

(b) Describe the behavior of g(x) =x2 − 5

x − 2near x = 2.

Solution:

(a) For f (x), we guess that

limx→−∞

f (x) = −1 and limx→∞

f (x) = 1.

The horizontal lines y = 1 and y = −1 are called “horizontalasymptotes” of the graph f (x).

(b) For g(x), we guess that

limx→2−

g(x) =∞ and limx→2+

g(x) = −∞.

The vertical line x = 2 is called a “vertical asymptote” ofthe graph g(x).

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Example 11 :




x2 + 1.


x − 2near x = 2.

Solution:


limx→−∞


f (x) = 1.



limx→2−


g(x) = −∞.


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Example 11 :




x2 + 1.


x − 2near x = 2.

Solution:


limx→−∞


f (x) = 1.



limx→2−


g(x) = −∞.


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Graphs of f (x) and g(x)

−4 −2 0 2 4

−1.

0−

0.5

0.0

0.5

1.0

f(x)

x

y

−4 −2 0 2 4

−10

010

20

g(x)

x

y

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Example 11 :

Reevaluate limx→∞



x2 + 1.

Solution: Note that

f (x) =x√

x2(1 + 1x2

)=

x

|x |√

(1 + 1x2

)=

sgn(x)√1 + 1

x2

.

We have

limx→∞

f (x) =limx→∞ sgn(x)

limx→∞

√1 + 1

x2

=1

1= 1.

Similarly, we have limx→−∞

f (x) = −1.

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Limits at infinity for rational functions

Corollary

Let Pm(x) = amxm + · · ·+ a0 and Qn(x) = bnx

n + · · ·+ b0 bepolynomials of degree m and n, respectively, so that am 6= 0 andbn 6= 0. Then

limx→−∞

Pm(x)

Qn(x)and lim

x→∞

Pm(x)

Qn(x)

(a) equals zero if m < n,

(b) equalsambn

if m = n,

(c) does not exist if m > n. More precisely, we say that the limitis ∞ or −∞, depending on the sign of the quotient as xbecomes large.

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Example 12 :


(a) limx→∞

4x2 − x + 3

3x2 + 5, (b) lim

x→∞

4x − 3

3x2 + 5, (c) lim

x→∞

4x2 − 3

3x + 5.

Solution:

(a) limx→∞

4x2 − x + 3

3x2 + 5= lim

x→∞

4− x−1 + 3x−2

3 + 5x−2=

4

3.

(b) limx→∞

4x − 3

3x2 + 5= lim

x→∞

4x−1 − 3x−2

3 + 5x−2=

0

3= 0.

(c) limx→∞

4x2 − 3

3x + 5= lim

x→∞

4x − 3x−1

3 + 5x−1=∞3

=∞.

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Example 12 :


(a) limx→∞

4x2 − x + 3

3x2 + 5, (b) lim

x→∞

4x − 3

3x2 + 5, (c) lim

x→∞

4x2 − 3

3x + 5.

Solution:

(a) limx→∞

4x2 − x + 3

3x2 + 5= lim

x→∞

4− x−1 + 3x−2

3 + 5x−2=

4

3.

(b) limx→∞

4x − 3

3x2 + 5= lim

x→∞

4x−1 − 3x−2

3 + 5x−2=

0

3= 0.

(c) limx→∞

4x2 − 3

3x + 5= lim

x→∞

4x − 3x−1

3 + 5x−1=∞3

=∞.

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Example 12 :


(a) limx→∞

4x2 − x + 3

3x2 + 5, (b) lim

x→∞

4x − 3

3x2 + 5, (c) lim

x→∞

4x2 − 3

3x + 5.

Solution:

(a) limx→∞

4x2 − x + 3

3x2 + 5= lim

x→∞

4− x−1 + 3x−2

3 + 5x−2=

4

3.

(b) limx→∞

4x − 3

3x2 + 5= lim

x→∞

4x−1 − 3x−2

3 + 5x−2=

0

3= 0.

(c) limx→∞

4x2 − 3

3x + 5= lim

x→∞

4x − 3x−1

3 + 5x−1=∞3

=∞.

54 / 75

Example 12 :


(a) limx→∞

4x2 − x + 3

3x2 + 5, (b) lim

x→∞

4x − 3

3x2 + 5, (c) lim

x→∞

4x2 − 3

3x + 5.

Solution:

(a) limx→∞

4x2 − x + 3

3x2 + 5= lim

x→∞

4− x−1 + 3x−2

3 + 5x−2=

4

3.

(b) limx→∞

4x − 3

3x2 + 5= lim

x→∞

4x−1 − 3x−2

3 + 5x−2=

0

3= 0.

(c) limx→∞

4x2 − 3

3x + 5= lim

x→∞

4x − 3x−1

3 + 5x−1=∞3

=∞.

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Example 13 :

Find the limit limx→∞

(√

x2 + x − x).

Solution: By rationalizing the expression we have

limx→∞

(√

x2 + x − x) = limx→∞

(√x2 + x − x)(

√x2 + x + x)√

x2 + x + x

= limx→∞

x√x2 + x + x

= limx→∞

1√1 + 1

x + 1

=1

2.

56 / 75

Example 14 :

Find the following limits if they exist:

(a) limx→2

x + 1

x − 2, (b) lim

x→1

x2 − 1

x2 − 3x + 2, (c) lim

x→1

√x − 1

x − 1.

Solution:

(a) limx→2

x + 1

x − 2does not exist because

limx→2−

x + 1

x − 2= −∞ and lim

x→2+

x + 1

x − 2=∞.

(b) limx→1

x2 − 1

x2 − 3x + 2= lim

x→1

x + 1

x − 2=

2

−1= −2.

(c) limx→1

√x − 1

x − 1= lim

x→1

1√x + 1

=1

2.

57 / 75

Example 14 :


(a) limx→2

x + 1

x − 2, (b) lim

x→1

x2 − 1

x2 − 3x + 2, (c) lim

x→1

√x − 1

x − 1.

Solution:

(a) limx→2

x + 1


limx→2−

x + 1


x→2+

x + 1

x − 2=∞.

(b) limx→1

x2 − 1

x2 − 3x + 2= lim

x→1

x + 1

x − 2=

2

−1= −2.

(c) limx→1

√x − 1

x − 1= lim

x→1

1√x + 1

=1

2.

58 / 75

Example 14 :


(a) limx→2

x + 1

x − 2, (b) lim

x→1

x2 − 1

x2 − 3x + 2, (c) lim

x→1

√x − 1

x − 1.

Solution:

(a) limx→2

x + 1


limx→2−

x + 1


x→2+

x + 1

x − 2=∞.

(b) limx→1

x2 − 1

x2 − 3x + 2= lim

x→1

x + 1

x − 2=

2

−1= −2.

(c) limx→1

√x − 1

x − 1= lim

x→1

1√x + 1

=1

2.

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Example 14 :


(a) limx→2

x + 1

x − 2, (b) lim

x→1

x2 − 1

x2 − 3x + 2, (c) lim

x→1

√x − 1

x − 1.

Solution:

(a) limx→2

x + 1


limx→2−

x + 1


x→2+

x + 1

x − 2=∞.

(b) limx→1

x2 − 1

x2 − 3x + 2= lim

x→1

x + 1

x − 2=

2

−1= −2.

(c) limx→1

√x − 1

x − 1= lim

x→1

1√x + 1

=1

2.

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§1.4 Continuity

Definition

We say that a function f is continuous at an interior point c of itsdomain if

limx→c

f (x) = f (c).

If either limx→c f (x) fails to exist or it exists but is not equal tof (c), then we will say that f is discontinuous at c .

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Example 15 :

Discuss the continuity of each of the following functions:

(a) f (x) =1

x, (b) g(x) =

x2 − 1

x + 1, (c) h(x) =

√x .

Solution:

(a) f (x) is continuous everywhere except for x = 0.

(b) g(x) is continuous everywhere except for x = −1.

(c) h(x) is continuous at any point in the domain of (0,∞).

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Example 15 :


(a) f (x) =1

x, (b) g(x) =

x2 − 1

x + 1, (c) h(x) =

√x .

Solution:




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Example 15 :


(a) f (x) =1

x, (b) g(x) =

x2 − 1

x + 1, (c) h(x) =

√x .

Solution:




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Example 15 :


(a) f (x) =1

x, (b) g(x) =

x2 − 1

x + 1, (c) h(x) =

√x .

Solution:




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Continuity of basic elementary functions

All polynomials are continuous everywhere.

sin(x), cos(x), arctan(x) and ex are continuous everywhere.

n√x is continuous for all x ∈ (−∞,∞) when n is odd, and for

x > 0 when n is even.

ln(x) is continuous on x ∈ (0,∞).

arcsin(x) and arccos(x) are continuous on x ∈ (−1, 1).

|x | is continuous everywhere.

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Properties of continuous functions

Assume that f (x) and g(x) are both continuous at point x = c .Then

1) kf (x) is continuous at c , where k is any number.

2) f (x)± g(x) is continuous at c .

3) f (x)g(x) is continuous at c .

4) f (x)/g(x) is continuous at c , if g(c) 6= 0.

5)√

f (x) is continuous at c , if f (c) > 0.

6) f ◦ g = f (g(x)) is continuous at c , if f (x) is also continuousat x = g(c).

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Exchanging limits and function evaluation

Theorem

If f (g(x)) is defined on an interval containing c , and if f iscontinuous at L and limx→c g(x) = L, then

limx→c

f (g(x)) = f (L) = f(

limx→c

g(x)).

Example: since cos is continuous everywhere, we have

limx→2

cos

(x2 − 4

x − 2

)= cos

(limx→2

x2 − 4

x − 2

)= cos(4).

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Example 16 :

Discuss the continuity of

h(x) = tan

(1

1− x

).

Solution:

h(x) can be written as h(x) = f (g(x)) with f (y) = tan(y),g(x) = 1

1−xg(x) is continuous at all points except at x = 1f (y) is continuous at all points, except where cos(y) = 0, i.e.,whenever y = (2k + 1)π2 , where k is an integerIf h is discontinuous at x = c , then either g discontinuous atx = c , or f is discontinuous at y = g(c)Thus, h is continuous everywhere, except at x = 1 and at

1

1− x= (2k + 1)

π

2⇐⇒ x = 1− 2

(2k + 1)π

for any integer k.69 / 75

Right and left continuity

Definition

(a) We say that f is right continuous at c if

limx→c+

f (x) = f (c).

(b) We say that f is left continuous at c if

limx→c−

f (x) = f (c).

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Continuity on an interval

Definition

(a) A function f (x) is said to be continuous on an open interval(a, b) if it is continuous at each point x = c in that interval.

(b) A function f (x) is said to be continuous on the closed interval[a, b] if it is continuous on (a, b) and

limx→a+

f (x) = f (a) and limx→b−

f (x) = f (b).

Or equivalently, ...... if it is continuous on (a, b), rightcontinuous at a, and left continuous at b.

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The Max-Min Theorem

Theorem

If f (x) is continuous on the closed, finite interval [a, b], then thereexist numbers p and q in [a, b] such that for all x in [a, b],

f (p) ≤ f (x) ≤ f (q).

Thus f has the absolute minimum value m = f (p), taken on at thepoint p, and the absolute maximum value M = f (q), taken on atthe point q.

Remark: The theorem merely asserts that the minimum andmaximum values exist; it does not tell us how to find them. Sometechniques for calculating the minimum and maximum values willbe introduced in Chapter 4.

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Intermediate Value Theorem

Theorem

[Intermediate Value Theorem] Suppose that f (x) is continuouson [a, b] and W is any number between f (a) and f (b). Then,there is at least one number c ∈ [a, b] for which

f (c) = W .

Remark: In other words, a continuous function attains all valuesbetween any two of its values. For instance, a girl who weighs 5pounds at birth and 100 pounds at age 12 must have weighedexactly 50 pounds at some time during her 12 years of life, sinceher weight is a continuous function of time.

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Intermediate Value Theorem

Corollary

Suppose that f (x) is continuous on [a, b] and f (a) and f (b) haveopposite signs [i.e., f (a)f (b) < 0]. Then, there is at least onenumber c ∈ (a, b) for which

f (c) = 0.

Remark: The intermediate value property has many applications.In particular, it can be used to estimate a solution of a givenequation.

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Example 17 :

Show that the equation x2 − x − 1 =1

x + 1has a solution between

1 < x < 2.

Proof: Let f (x) = x2 − x − 1− 1

x + 1. Then f (1) = −3/2 and

f (2) = 2/3. Since f (x) is continuous for 1 ≤ x ≤ 2, it follows fromthe intermediate value theorem that the curve must cross the xaxis somewhere between x = 1 and x = 2.

In other words, there is a number c such that 1 < c < 2 andf (c) = 0, i.e.,

c2 − c − 1 =1

c + 1.

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Chapter 1: Limits and Continuityfelix_kwok/2018/chapter1_notes1.pdf · Chapter 1: Limits and Continuity Spring 2018 Department of Mathematics Hong Kong Baptist University 1/75. x1.1