lecture 11: continuity

Lecture 11: Continuity

Objective:

� Present the concept of continuity (at a point, on an interval, and from one-side)

� Discuss the properties of continuous functions and introduce The Intermediate ValueTheorem.

� Suggested problems: Section 2.5 #1 − 63 (odd)

Introduction

The word continuous occurs often in common parlance and refers to anything that has nobreaks or gaps in it. Similarly, in mathematics, a continuous function is a real valued functionthat has no holes, no gaps, no breaks, and no jumps in it. Loosely speaking, a continuousfunction is one whose graph you can draw without lifting your pen from the paper.

In the study of calculus, the continuity of a function is a necessary condition for differentiation,and a sufficient condition for integration.

Continuity

The concept of continuity is simple: If the graph of the function doesn’t have any breaks orholes in it within a certain interval, the function is said to be continuous over that interval.

Drawing the graph of a function is one way to ascertain if a function is continuous or not - butthis is not always feasible. To overcome these issues, a technical definition is needed. There arethree types continuity: continuity at a single point, continuity from one-side, and continuity onan interval.

1

2

Continuity at a Single Point

Definition 1

A function f is continuous at the point x = a if and only if

limx→a

f(x) = f(a)

The animation below illustrates that in order for a function to be continuous at the point x = a,not only do the left and right limits have to agree as x → a (implying that the general limitexists), but that the function’s value at a, must be equal to the limit of the function at x = a.That is, lim

x→af(x) = f(a)

Example 1

Use the definition of continuity and the property of limits to show that the function iscontinuous at the given number a

f(x) = x3 − x2 − 2x a = 5

Solution 1

We are given f(x) = x3 − x2 − 2x

To show that f(x) is continuous at a = 5 we need to show that limx→5

f(x) = f(5).

limx→5

f(x) = limx→5

(x3 − x2 − 2x) = limx→5

x3 − limx→5

x2 − 2 · limx→5

x = (5)3 − (5)2 − 2(5) = 90

= f(5)

∴ By the definition of continuity, the function f is continuous at a = 5

3

Example 2


g(t) =t2 + 5t

2t+ 1a = 2

Solution 2

We are given g(t) =t2 + 5t

2t+ 1

To show that g(t) is continuous at a = 2 we need to show that limt→2

g(t) = g(2).

limt→2

g(t) = limt→2

t2 + 5t

2t+ 1=

limt→2

(t2 + 5t)

limt→2

(2t+ 1)=

limt→2

t2 + 5 · limt→2

t

2 · limt→2

t+ 1=

(2)2 + 5(2)

2(2) + 1=

14

5= g(2)

∴ By the definition of continuity, the function g is continuous at a = 2

Example 3


f(r) =3√4r2 − 2r + 7 a = −2

Solution 3

We are given f(r) = 3√4r2 − 2t+ 7

To show that f(r) is continuous at a = −2 we need to show that limr→−2

f(r) = f(−2).

limr→−2

f(r) = limr→−2

3√4r2 − 2r + 7 = 3

√limr→−2

(4r2 − 2r + 7) = 3

√4 · lim

r→−2r2 − 2 · lim

r→−2r + 7

= 3√

4(−2)2 − 2(−2) + 7

=3√27

= f(−2)

∴ By the definition of continuity, the function f is continuous at a = −2

4

Example 4

Consider the function

f(x) =

1

x+ 2; x ̸= −2

1 ; x = −2

Explain why the function is discontinuous at a = −2. Sketch the graph of the function.

Solution 4

Starting from the definition of continuity, observe that f(−2) = 1

But

limx→−2−

1

x+ 2= −∞ and lim

x→−2+

1

x+ 2= +∞

∵ limx→−2

f(x) = dne ⇒ f(x) is discontinuous at a = −2

(−2, 1)

x

y

Example 5


f(x) =

x2 − x

x2 − 1; x ̸= 1

1 ; x = 1

Explain why the function is discontinuous at a = 1. Sketch the graph of the function.

Solution 5

Starting from the definition of continuity, observe that f(1) = 1

5

But

limx→1

x2 − x

x2 − 1= lim

x→1

x(x− 1)

(x− 1)(x+ 1)= lim

x→1

x��(x− 1)

��(x− 1)(x+ 1)= lim

x→1

x

x+ 1=

1

2

∵ limx→1

f(x) ̸= f(1) ⇒ f(x) is discontinuous at a = 1

1

x

y

Example 6


f(x) =

2x2 − 5x− 3

x− 3; x ̸= 3

6 ; x = 3

Explain why the function is discontinuous at a = 3. Sketch the graph of the function.

Solution 6

Starting from the definition of continuity, observe that f(3) = 6

But

limx→3

2x2 − 5x− 3

x− 3= lim

x→3

(2x+ 1)(x− 3)

x− 3= lim

x→3

(2x+ 1)��(x− 3)

��x− 1= lim

x→3(2x+ 1) = 7

∵ limx→3

f(x) ̸= f(3) ⇒ f(x) is discontinuous at a = 3

3

76

x

y

6

Types of Discontinuity

There are three types of discontinuities: removable, infinite, and jump discontinuities. Thegraphs below illustrate the three different types.

ax

y

ax

y

ax

y

ax

y

Removable Removable Infinite Jumplimx→a

f(x) ̸= f(a) limx→a

f(x) exists limx→a

f(x) = ∞ limx→a−

f(x) ̸= limx→a+

f(x)

but f(a) = und

One-Sided Continuity

Similar to one-sided limits, we can also define one-sided continuity.

Definition 2

A function f is continuous from the right of a if limx→a+

f(x) = f(a)

and f is continuous from the left of a if limx→a−

f(x) = f(a)

Example 7

The graph of f is shown below.

124 CHAPTER 2 Limits and Derivatives

2.5 Exercises

1. Write an equation that expresses the fact that a function f is continuous at the number 4.

2. If f is continuous on s2`, `d, what can you say about its graph?

3. (a) From the given graph of f , state the numbers at which f is discontinuous and explain why.

(b) For each of the numbers stated in part (a), determine whether f is continuous from the right, or from the left, or neither.

y

x_4 2 4 6_2 0

4. From the given graph of t, state the numbers at which t is discontinuous and explain why.

10_3 _2 2 3

y

x

5– 6 The graph of a function f is given.(a) At what numbers a does lim x l a f sxd not exist?(b) At what numbers a is f not continuous?(c) At what numbers a does lim x l a f sxd exist but f is not

continuous at a ?

5. y

0 x1

6. y

0 x1

7 – 10 Sketch the graph of a function f that is defined on R and continuous except for the stated discontinuities.

7. Removable discontinuity at 22, infinite discontinuity at 2

8. Jump discontinuity at 23, removable discontinuity at 4

9. Discontinuities at 0 and 3, but continuous from the right at 0 and from the left at 3

10. Continuous only from the left at 21, not continuous from the left or right at 3

11. The toll T charged for driving on a certain stretch of a toll road is $5 except during rush hours (between 7 am and 10 am and between 4 pm and 7 pm) when the toll is $7.

(a) Sketch a graph of T as a function of the time t, measured in hours past midnight.

(b) Discuss the discontinuities of this function and their significance to someone who uses the road.

12. Explain why each function is continuous or discontinuous. (a) The temperature at a specific location as a function of

time (b) The temperature at a specific time as a function of the

distance due west from New York City (c) The altitude above sea level as a function of the distance

due west from New York City (d) The cost of a taxi ride as a function of the distance

traveled (e) The current in the circuit for the lights in a room as a

function of time

13–16 Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.

13. f sxd − 3x 2 1 sx 1 2d5, a − 21

14. tstd −t 2 1 5t

2t 1 1 , a − 2

15. psvd − 2s3v2 1 1 , a − 1

16. f srd − s3 4r 2 2 2r 1 7 , a − 22

17 –18 Use the definition of continuity and the properties of limits to show that the function is continuous on the given interval.

17. f sxd − x 1 sx 2 4 , f4, `d

18. tsxd −x 2 1

3x 1 6, s2`, 22d

19– 24 Explain why the function is discontinuous at the given number a. Sketch the graph of the function.

19. f sxd −1

x 1 2 a − 22

20. f sxd − H 1

x 1 2

1

if x ± 22

if x − 22

a − 22

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Assume that the domain of f is [−3, 3]. State the numbers at which f is discontinuous.Categorize each as removable, jump, or infinite discontinuity, and if it is continuous fromthe left of a, the right of a, or neither.

7

Solution 7

a = −2: Removable discontinuity ∵ f(−2) = und.Neither left nor right continuous ∵ lim

x→−2−f(x) ̸= f(−2) and lim

x→−2+f(x) ̸= f(−2).

a = −1: Infinite discontinuity. ∵ limx→−1

f(x) = −∞.

Neither left nor right continuous ∵ limx→−1−

f(x) ̸= f(−1) and limx→−1+

f(x) ̸= f(−1).

a = 0: Jump discontinuity. ∵ limx→0

f(x) = dne

Continuous from the left of a ∵ limx→0−

f(x) = f(0).

a = 1: Jump discontinuity. ∵ limx→1

f(x) = dne

Neither left nor right continuous ∵ limx→1−

f(x) ̸= f(1) and limx→1+

f(x) ̸= f(1).

Example 8


f(x) =

2x ; x ≤ 1

3− x ; 1 < x ≤ 4√x ; x > 4

Find the numbers at which f is discontinuous. At which of these numbers is f continuousfrom the left, from the right, or neither? Sketch the graph of f .

Solution 8

The function f has potential discontinuities at a = 1 and at a = 4

At a = 1, we have

limx→1−

f(x) = limx→1−

2x = 2

limx→1+

f(x) = limx→1+

(3− x) = 3− 1 = 2

f(1) = 2

∵ limx→1

f(x) = f(1) ⇒ f is continuous at a = 1

At a = 4, we have

limx→4−

f(x) = limx→4−

(3− x) = 3− 4 = −1

limx→4+

f(x) = limx→4+

√x =

√4 = 2

8

f(4) = −1

∵ limx→4−

f(x) = f(4) ⇒ f is left-continuous at a = 4

Below is the graph of f(x)

1 4

2

−1

x

y

Example 9


f(x) =

x+ 1 ; x ≤ 11

x; 1 < x < 3

√x− 3 ; x ≥ 3

Find the numbers at which f is discontinuous. At which of these numbers is f continuousfrom the left, from the right, or neither? Sketch the graph of f .

Solution 9

The function f has potential discontinuities at a = 1 and at a = 3

At a = 1, we have

limx→1−

f(x) = limx→1−

(x+ 1) = 2

limx→1+

f(x) = limx→1+

1

x= 1

f(1) = 2

∵ limx→1−

f(x) = f(1) ⇒ f is continuous from the left of a = 1

At a = 3, we have

limx→3−

f(x) = limx→3−

1

x=

1

3limx→3+

f(x) = limx→3+

√x− 3 =

√3− 3 = 0

9

f(3) = 0

∵ limx→3+

f(x) = f(3) ⇒ f is continuous from the right of a = 3

Below is the graph of f(x)

1 3

2

1

1/3x

y

Continuity on an Interval

Definition 3

A function f is continuous on an interval if it is continuous at every number in theinterval. If f is defined only on one side of an endpoint of the interval then continuous atthe endpoint is understood to be a one-sided continuity.

Example 10

Use the definition of continuity and the property of limits to show that the function iscontinuous on the given interval

h(x) = 1−√4− x2 [−2, 2]

Solution 10

The easiest way to show that the function is continuous everywhere in [−2, 2] is to firstshow that lim

x→af(x) = f(a) whenever a ∈ (−2, 2), and then to show that the function is

right continuous at the endpoint a = −2, and left-continuous on a = 2.

For −2 < a < 2 we have that

limx→a

f(x) = limx→a

(1−√4− x2)

= limx→a

1−√

limx→a

4− limx→a

x2

= 1−√4− a2 ; but this is precisely f(a)

10

∵ limx→a

f(x) = f(a) ⇒ f(x) is continuous at x = a ∀a ∈ (−2, 2)

Now for the endpoints.

At a = −2 we just need to check that it is continuous to the right of a = −2, becausethe function is defined only to the right of it.

limx→−2+

f(x) = limx→−2+

1−√4− x2 = 1−

√4− (−2)2 = 1− 0 = f(−2)

At a = +2 we just need to check that it is continuous to the left of a = +2, becausethe function is defined only to the left of it.

limx→+2−

f(x) = limx→+2−

1−√4− x2 = 1−

√4− (2)2 = 1− 0 = f(2)

∵ The function is continuous at the endpoints, and everywhere in between, f(x) is contin-uous on [−2, 2]

Here is the graph of f(x) = 1−√4− x2

−2 2

1

−1

x

y

Example 11


f(x) =2x+ 3

x− 2(2,∞)

Solution 11

For 2 < a we have that

limx→a

f(x) = limx→a

2x+ 3

x− 2=

limx→a

(2x+ 3)

limx→a

(x− 2)=

2 · limx→a

x+ limx→a

3

limx→a

x− limx→a

2=

2a+ 3

a− 2= f(a)

∵ limx→a

f(x) = f(a) ⇒ f(x) is continuous at x = a ∀a ∈ (2,∞)

11

We do not need to check that a = 2 is continuous from the right, because the endpoint isexcluded (i.e. this is an open interval).

2

2

f(x) =2x+ 3

x− 2

x

y

Example 12


g(x) = 2√3− x (−∞, 3]

Solution 12

For a < 3 we have that

limx→a

f(x) = limx→a

2√3− x = 2

√limx→a

3− limx→a

x = 2√3− a = f(a)

∵ limx→a

f(x) = f(a) ⇒ f(x) is continuous at x = a ∀a ∈ (−∞, 3)

Now for the endpoint.

At a = 3 we just need to check that it is continuous from the left, becausethe function is defined only to the left of it.

limx→3−

f(x) = limx→3−

2√3− x = 2

√3− 3 = 0 = f(3)

∵ The function is continuous at the endpoint, and at all values to the left of it, f(x) =2√3− x is continuous on (−∞, 3]

3

f(x) = 2√3− x

x

y

12

Properties of Continuous Functions

Continuity is an important property of because of how it interacts with other properties offunctions. Functions which are continuous are in some sense ”the nicest functions” becausemany proofs in real analysis rely on approximating arbitrary functions with continuous ones.

The following theorems show that the basic properties of continuity follow from those of limits.

Theorem 1

If f and g are continuous at a and c is a constant, then the following functions are alsocontinuous at a

f + g f − g cf fgf

g; g(a) ̸= 0

Theorem 2

The following types of functions are continuous at every number in their domains

� Polynomials

� Rational functions

� Root functions

� Trigonometric functions

� Inverse trigonometric functions

� Exponential functions

� Logarithmic functions

Theorem 3

If f is continuous at b and limx→a

g(x) = b then limx→a

f(g(x)) = f(b) That is

limx→a

f(g(x)) = f[limx→a

g(x)]= f(b)

Theorem 4

If g is continuous at a and f is continuous at g(a), then the composite function

f ◦ g = (f ◦ g)(x) = f(g(x))

is also continuous at a.

13

Example 13

For each of the following functions, explain using Theorems 1 − 4 why the function iscontinuous at every number in its domain. State the domain of the function.

a. g(v) =3v − 1

v2 + 2v − 15

b. B(u) =√3u− 2 + 3

√2u− 3

c. f(t) = e−t2 ln(1 + t2)

d. g(t) = tan−1(et − 1)

Solution 13

a. g(v) =3v − 1

v2 + 2v − 15

g(v) is a rational function. Rational functions are continuous on its domain (Theorem2).

g(v) =3v − 1

v2 + 2v − 15=

3v − 1

(v + 5)(v − 3)

The domain of g is {v | v ̸= −5, 3} = (−∞,−5) ∪ (−5, 3) ∪ (3,∞)

b. B(u) =√3u− 2 + 3

√2u− 3

B(u) is the sum of two root functions. Let g(u) =√3u− 2 and h(u) = 3

√2u− 3

Root functions are continuous on their domains (Theorem 2) and the sum of twocontinuous functions is also a continuous function (Theorem 1).

∵ DB = Dg ∩Dh =[23,∞

)∩ (−∞,∞) ⇒ The domain of B is

[23,∞

)c. f(t) = e−t2 ln(1 + t2)

f(t) is the product of two functions. Let g(t) = e−t2 and h(t) = ln(1 + t2).

Exponential and logarithmic functions are continuous on their domains (Theorem 2)and the product of two continuous functions is also a continuous function (Theorem1). Moreover, each component of the product, g(t) and h(t), are composite functions.For the function g(t), −t2 is continuous on (−∞,∞) and so is e−t2 (Theorem 3). Forthe function h(t), t2 +1 is continuous on (−∞,∞), and so is ln(t2 +1) (Theorem 4).

The domain of f(t) = (−∞,∞)

d. g(t) = tan−1(et − 1)

g(t) is a composition of two functions: an inverse trigonometric function and a poly-nomial. Both are continuous on their domains (Theorem 2). Let f(t) = tan−1(t) andh(t) = et−1, then g(t) = (f◦h)(t) = f(h(t)). Then by Theorem 4, g(t) = tan−1(et−1)is continuous on (−∞,∞) ∵ Dh = (−∞,∞) and Df = (−∞,∞)

14

Example 14

Use continuity to evaluate the limit.

a. limx→π

sin(x+ sinx)

b. limx→4

3√x2−2x−4

c. limx→1

ln

(5− x2

1 + x

)

Solution 14

a. limx→π

sin(x+ sinx)

∵ x is continuous on R and x+sinx is also continuous on R, the composite function,f(x) = sin(x+ sinx) is also continuous on R. ∵ π ∈ Df ⇒ lim

x→πf(x) = f(π)

limx→π

sin(x+ sinx) = f(π) = sin(π + sin(π)) = sin(π + 0) = 0

b. limx→4

3√x2−2x−4

The function f(x) = 3√x2−2x−4 is continuous on its domain, since it is a composition

of an exponential function, a root function, and a polynomial. Using a sign chart orotherwise, we find that the domain of f(x) is Df = (−∞, 1−

√5]∪ [1 +

√5,∞), and

∵ 4 ∈ Df limx→4

f(x) = f(4)

limx→4

3√x2−2x−4 = f(4) = 3

√42−2(4)−4 = 32 = 9

c. limx→1

ln

(5− x2

1 + x

)The function f(x) = ln

(5−x2

1+x

)is continuous on its domain, since it is a composition

of a logarithmic function and a rational function. Using a sign chart or otherwise,we find that the domain of f(x) is Df = (−∞,−

√5] ∪ [

√5,∞), and ∵ 1 ∈ Df ⇒

limx→1

f(x) = f(1)

limx→1

ln

(5− x2

1 + x

)= f(1) = ln

(5− 1

1 + 1

)= ln 2

15

Example 15

For what value of k is f continuous on (−∞,∞)?

f(x) =

{kx2 ; x ≤ −6

5x+ k ; x > −6

Solution 15

In order for f(x) to be continuous on (−∞,∞), we must ensure that limx→a

f(x) = f(a)

There is only value that f could potentially be discontinuous at is a = −6

Since the general limit must exist, this implies that

limx→−6−

f(x) = limx→−6+

f(x)

limx→−6−

kx2 = limx→−6+

(5x+ k)

k(−6)2 = 5(−6) + k

36k = −30 + k

k = −6

7

Therefore,

f(x) =

−6

7x2 ; x ≤ −6

5x− 6

7; x > −6

16

Example 16

Find the values of c and d which make f continuous everywhere.

f(x) =

x2 − 4

x− 2; x < 2

cx2 − dx+ 3 ; 2 ≤ x < 3

2x− c+ d ; x ≥ 3

Solution 16

There are two values of x where the function f may be discontinuous: a = 2 and a = 3

a = 2

limx→2−

f(x) = limx→2+

f(x)

limx→2−

x2 − 4

x− 2= lim

x→2+(cx2 − dx+ 3)

limx→2−

��(x− 2)(x+ 2)

��x− 2= lim

x→2+(cx2 − dx+ 3)

limx→2−

(x+ 2) = limx→2+

(cx2 − dx+ 3)

4 = 4c− 2d+ 3

4c− 2d = 1

a = 3

limx→3−

f(x) = limx→3+

f(x)

limx→3−

(cx2 − dx+ 3) = limx→3+

(2x− c+ d)

9c− 3d+ 3 = 6− c+ d

10c− 4d = 3

Using the two equations: 4c − 2d = 1 and 10c − 4d = 3 we can solve for c and d, we findthat c = 1

2and d = 1

2. Thus,

f(x) =

x2 − 4

x− 2; x < 2

12x2 − 1

2x+ 3 ; 2 ≤ x < 3

2x− 12+ 1

2; x ≥ 3

17

The Intermediate Value Theorem (IVT)

The Intermediate Value Theorem is a precise mathematical statement concerning the propertyof a continuous function.

The Intermediate Value Theorem

Suppose that f is a continuous function on the closed interval [a, b], and let N be anynumber between f(a) and f(b) such that f(a) ̸= f(b). Then there exists a number c in(a, b) such that f(c) = N .

122 CHAPTER 2 Limits and Derivatives

EXAMPLE 9 Where are the following functions continuous?(a) hsxd − sinsx 2 d (b) Fsxd − lns1 1 cos xd

SOLUTION (a) We have hsxd − f stsxdd, where

tsxd − x 2 and f sxd − sin x

We know that t is continuous on R since it is a polynomial, and f is also continuous everywhere. Thus h − f 8 t is continuous on R by Theorem 9.

(b) We know from Theorem 7 that f sxd − ln x is continuous and tsxd − 1 1 cos x is continuous (because both y − 1 and y − cos x are continuous). Therefore, by Theo-rem 9, Fsxd − f stsxdd is continuous wherever it is defined. The expression ln s1 1 cos xd is defined when 1 1 cos x . 0, so it is undefined when cos x − 21, and this happens when x − 6�, 63�, . . . . Thus F has discontinuities when x is an odd multiple of � and is continuous on the intervals between these values (see Figure 7). ■

■ The Intermediate Value TheoremAn important property of continuous functions is expressed by the following theorem, whose proof is found in more advanced books on calculus.

10 The Intermediate Value Theorem Suppose that f is continuous on the closed interval fa, bg and let N be any number between f sad and f sbd, where f sad ± f sbd. Then there exists a number c in sa, bd such that f scd − N.

The Intermediate Value Theorem states that a continuous function takes on every inter-mediate value between the function values f sad and f sbd. It is illustrated by Figure 8. Note that the value N can be taken on once [as in part (a)] or more than once [as in part (b)].

(b)

0 x

y

f(b)

N

f(a)

a c£ b

y=ƒ

c™c¡

(a)

0 x

y

f(b)

N

f(a)

b

y=ƒ

a c

FIGURE 8

If we think of a continuous function as a function whose graph has no hole or break, then it is easy to believe that the Intermediate Value Theorem is true. In geometric terms it says that if any horizontal line y − N is given between y − f sad and y − f sbd as in Fig-ure 9, then the graph of f can’t jump over the line. It must intersect y − N somewhere.

It is important that the function f in Theorem 10 be continuous. The Intermediate Value Theorem is not true in general for discontinuous functions (see Exercise 52).

2

_6

_10 10

FIGURE 7 y − lns1 1 cos xd

b0 x

y

f(a)

N

f(b)

a

y=N

y=ƒ

FIGURE 9

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Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-200-203

In other words, the IVT guarantees that if a continuous function attains two different values,then it must also attain all values in between these two values. This is particularly useful ifwe want to show that a function attains a certain value, or that an equation has a particularsolution.

Example 17

Use the Intermediate Value Theorem to show that there is a solution to the given equationin the specified interval.

lnx = x−√x (2, 3)

Solution 17

To show that ln x = x−√x has a solution in (2, 3), we want to do is rewrite the equation

as follows:

lnx− x+√x = 0

because solving for when this equation equals 0 is a lot easier than determining whenlnx = x−

√x.

Let f(x) = lnx − x +√x. The domain of f is (0,∞), so it is also continuous on (2, 3).

Now we pick a value of x in the interval (2, 3) and ascertain the sign of the result

f(2.2) = ln(2.2)− 2.2 +√2.2 = 0.0717 > 0

Now pick another value in the interval (2, 3) so that f(x) < 0

18

f(2.9) = ln(2.9)− 2.9 +√2.9 = −0.1323 < 0

Since f(2.2) > 0 and f(2.9) < 0 and f(x) in continuous on (2, 3) ⇒ ∃c ∈ (2, 3) such thatf(x) = 0 (which implies that ln x− x+

√x = 0 has a solution in the given interval).

2 3x

y

Example 18

Use the Intermediate Value Theorem to show that there is a solution to the given equationin the specified interval.

sinx = x2 − x (1, 2)

Solution 18

To show that sinx = x2 − x has a solution in (1, 2) consider the equation

sinx− x2 + x = 0

Let f(x) = sinx− x2 + x. The domain of f is (−∞,∞), so it is also continuous on (1, 2).

∵ f(1.5) > 0 and f(1.9) < 0 and f(x) in continuous on (1, 2) ⇒ ∃c ∈ (1, 2) such thatf(x) = 0 (which implies that sin x = x2 − x has a solution in the given interval).

1 2x

y

19

Example 19

If f and g are continuous functions at a, prove that f + g is also continuous at a

Solution 19

Let f and g be two continuous functions at a. Then limx→a

f(x) = f(a) and limx→a

g(x) = g(a).

As a result,

limx→a

[f(x) + g(x)] = limx→a

f(x) + limx→a

g(x)

= f(a) + g(a)

= [f + g](a)

Example 20

If f and g are continuous functions at a, prove that f − g is also continuous at a

Solution 20

Suppose that f and g are two continuous functions at a. Then limx→a

f(x) = f(a) and

limx→a

g(x) = g(a). As a result,

limx→a

[f(x)− g(x)] = limx→a

f(x)− limx→a

g(x)

= f(a)− g(a)

= [f − g](a)

Example 21

If f and g are continuous functions at a, prove that fg is also continuous at a

Solution 21



g(x) = g(a).

20

As a result,

limx→a

[f(x)g(x)] = limx→a

f(x) · limx→a

g(x) = f(a) · g(a) = [fg](a)

Example 22

If f and g are continuous functions at a, prove thatf

gis also continuous at a; provided

that g(a) ̸= 0.

Solution 22



g(x) = g(a).

As a result,

limx→a

[f(x)

g(x)

]=

limx→a

f(x)

limx→a

g(x)=

f(a)

g(a)=

[f

g

](a)

Example 23

If f and g are continuous functions at a, prove that cf is also continuous at a

Solution 23

Let f is a continuous functions at a, then limx→a

f(x) = f(a) and c is a real constant. Then,

limx→a

[c · f(x)] = c · limx→a

f(x) = c · f(a) = [cf ](a)

lecture 11: continuity

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