chapter 2 limits and continuitymathstat.sci.tu.ac.th/~archara/ma111/ma111-118/... · limits and...

Chapter 2

Limits and Continuity

2.1 Limits

The limits is the fundamental notion of calculus. This underlying concept is the threadthat binds together virtually all of the calculus you are about to study.

In this section, we develop the notion of limit using some common language andillustrate the idea with some simple examples.

As a start, consider the function

f(x) =x2 − 1

x− 1and g(x) =

x2 − 2

x− 1

Notice that both functions are undefined at x = 1. But we can examine their behaviorfor x close to 1, as in the following table.

x f(x) =x2 − 1

x− 1g(x) =

x2 − 2

x− 10.9 1.9 11.90.99 1.99 101.990.999 1.999 1, 001.9990.9999 1.9999 10, 001.99990.99999 1.99999 100, 001.999990.999999 1.999999 10, 000, 001.999999

Notice that as you move down the first column of the table, the x-values get closerto 1, but all less than 1. We use the notation x → 1− to indicate that x approaches

1 from the left side. Notice that the table suggest that as x gets closer and closer to1 (with x < 1), f(x) is getting closer and closer to 2. In view of this, we say that thelimit of f(x) as x approaches 1 from the left is 2, written

limx→1−

f(x) = 2.

On the other hand, the table indicate that as x gets closer and closer to 1 (with x < 1),g(x) increases without bound. In this case, g(x) is said to increase without bound asx approaches 1 from the left, written

limx→1−

g(x) = +∞

3

MA111: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 4

Likewise, we need to consider what happens to the function values for x close to 1but larger than 1. Here, we use the notation x → 1+ to indicate that x approaches 1from the right side. We compute some of these values in the following table.

x f(x) =x2 − 1

x− 1g(x) =

x2 − 2

x− 11.1 2.1 −7.91.01 2.01 −97.991.001 2.001 −997.9991.0001 2.0001 −9, 997.99991.00001 2.00001 −99, 997.999991.000001 2.000001 −999, 997.999999

Again, the table suggest that as x gets closer and closer to 1 (with x > 1), f(x)is getting closer and closer to 2. In view of this, we say that the limit of f(x) as xapproaches 1 from the right is 2, written

limx→1+

f(x) = 2.

Finally, the table suggest that g(x) decreases without bound as x approaches 1 fromthe right. We say that

limx→1+

g(x) = −∞

We call limx→1−

f(x) and limx→1+

f(x) one-sided limits. Since the two one-sided limits

of f(x) are the same, we summarize our results by saying that the limit of f(x) as xapproaches 1 is 2, written

limx→1

f(x) = 2

On the other hand, g(x) do not approach a real number as x approaches 1 either fromthe left or from the right, and

limx→1−

g(x) = +∞ and limx→1+

g(x) = −∞.

In this case, we say thatlimx→1

g(x) does not exist

Hence a limit exists if and only if both one-sided limits exist and are equal.That is,

limx→a

f(x) = L, for some number L, if and only if limx→a−

f(x) = limx→a+

f(x) = L

Moreover, we can write

limx→1

f(x) = limx→1

x2 − 1

x− 1= lim

x→1

(x− 1)(x+ 1)

x− 1= lim

x→1(x+ 1) = 2

where we can cancel the factors (x− 1) since in the limit as x → 1, x is close to 1, butx 6= 1, so that x− 1 6= 0.


Determine whether or not limx→0

1

xexists.

Example 2.1.

Solution We first graph y =1

xand compute some function values for x close to 0.

−3 3

10

−10

x

y

y =1

x

x 1/x±0.1 ±10±0.01 ±100±0.001 ±1000±0.0001 ±10, 000±0.00001 ±100, 000

Notice that, as x → 0+,1

xincreases without bound. Thus,

limx→0+

1

x= +∞.

Likewise, we can say that

limx→0−

1

x= −∞.

Therefore,

limx→0

1

xdoes not exist. z

Evaluate limx→0

sin x

x.

Example 2.2.

Solution We graph f(x) =sin x

xand compute some function values.

x

y

y =sinx

x

1

0 1 2 3−1−2−3

x (sin x)/x

±0.1 0.998334±0.01 0.999983±0.001 0.99999983±0.0001 0.9999999983±0.00001 0.999999999983


The graph and the tables of value lead us to

limx→0+

sin x

x= 1 and lim

x→0−

sin x

x= 1

Thus,

limx→0

sin x

x= 1. z

1

2

3

4

5

6

−1

−2

−3

1 2 3 4 5 6 7−1−2−3−4

x

y

b bc

b

bc

For the function whose graph is given, state the value of the given quantity, ifit does not exist, explain why.

(a) limx→2−

f(x) = (b) limx→2+

f(x) =

(c) limx→2

f(x) = (d) limx→−1−

f(x) =

(e) limx→−1+

f(x) = (f) limx→−1

f(x) =

Example 2.3.


Exercises 2.1

1.

x

y

1

y = f(x)

For the function whose graph is given, state the value of the given quantity, if itdoes not exist, explain why.

(a) limx→0−

f(x) (b) limx→0+

f(x)

(c) limx→0

f(x) (d) limx→2−

f(x)

(e) limx→2+

f(x) (f) limx→2

f(x)

(g) limx→−3

f(x)

2.

-4 -3 -2 -1 0 1 2 3 4 5

-4

-3

-2

-1

0

1

2

3

4 y

x

b

b

y = f(x)


(a) limx→1

f(x) (b) limx→3−

f(x)

(c) limx→3+

f(x) (d) limx→3

f(x)

(e) f(3) (f) limx→−2−

f(x)

(g) limx→−2+

f(x) (h) limx→−2

f(x)

(i) f(−2)


3.

-4 -3 -2 -1 0 1 2 3 4-1

0

1

2

3

4

x

y

bcbc

b

b

b

bc

y = f(x)


(a) limx→−3

f(x) (b) f(−3)

(c) f(−1) (d) limx→−1

f(x)

(e) f(1) (f) limx→1−

f(x)

(g) limx→1+

f(x) (h) limx→1

f(x)

Answer to Exercises 2.1

1. (a) −2 (b) 2 (c) Does not exist (d) −1 (e) 3 (f) Does not exist (g) 2

2. (a) 3 (b) 2 (c) −2 (d) Does not exist (e) 1 (f) −1 (g) −1 (h) −1 (i) −3

3. (a) 2 (b) 1 (c) 2 (d) 5

2(e) 2 (f) 2 (g) 1 (h) Does not exist

2.2 Computing Limits

In this section we will discuss techniques for computing limits of many functions. Westart with the following basic results.

Let a and k be real numbers.

(a) limx→a

k = k (b) limx→a

x = a (c) limx→0−

1

x= −∞ (d) lim

x→0+

1

x= +∞

Theorem 2.1.

The following theorem will be our basic tool for finding limits algebraically.


Let a and k be real numbers. Suppose that limx→a

f(x) = L1 and limx→a

g(x) = L2

Then:

(a) limx→a

[

kf(x)]

= k limx→a

f(x) = kL1

(b) limx→a

[

f(x) + g(x)]

= limx→a

f(x) + limx→a

g(x) = L1 + L2

(c) limx→a

[

f(x)− g(x)]

= limx→a

f(x)− limx→a

g(x) = L1 − L2

(d) limx→a

[

f(x) · g(x)]

= limx→a

f(x) · limx→a

g(x) = L1L2

(e) limx→a

f(x)

g(x)=

limx→a

f(x)

limx→a

g(x)=

L1

L2

, provided L2 6= 0

(f) limx→a

[

f(x)]n

=[

limx→a

f(x)]n

= Ln1 , n a positive integer

(g) limx→a

n

√

f(x) = n

√

limx→a

f(x) = n

√

L1, provided L1 > 0 if n is even.

Moreover, these statements are also true for the one-sided limits as x → a− oras x → a+.

Theorem 2.2 (Limit Laws).

Apply the rules of limits to evaluate the following.

(a) limx→2

(x3 + 4x2 − 3) (b) limx→3

x4 + x2 − 1

x2 + 5(c) lim

x→5

3√3x2 − 4x+ 9

Example 2.4.

Solution


For any polynomial

p(x) = cnxn + cn−1x

n−1 + · · ·+ c1x+ c0

and any real number a,

limx→a

p(x) = cnan + cn−1a

n−1 + · · ·+ c1a+ c0 = p(a)

Theorem 2.3 (Limits of Polynomials).

Let f(x) =p(x)

q(x)be the rational function, and let a be any real number.

(a) If q(a) 6= 0, then limx→a

f(x) = f(a).

(b) If q(a) = 0 but p(a) 6= 0, then limx→a

f(x) does not exist.

Theorem 2.4 (Limits of Rational Functions).

Evaluate limx→2

5x3 + 2x− 5

x2 − 3.

Example 2.5.

Solution

Evaluate limx→1

3x2 − x− 2

2x2 + x− 3.

Example 2.6.

Solution


Evaluate limx→1

(

1

x− 1− 2

x2 − 1

)

.

Example 2.7.

Solution

Evaluate limx→−3−

|x+ 3|x3 + x2 − 6x

.

Example 2.8.

Solution

Evaluate limx→2

x2/3 − 41/3

x− 2.

Example 2.9.

Solution


Evaluate limx→4

x2 − 5x+ 4√x− 2

.

Example 2.10.

Solution

Evaluate limx→3

√x+ 1− 2

x3 − 27.

Example 2.11.

Solution


Evaluate limx→2

2− 3√x+ 6

x− 2.

Example 2.12.

Solution

For functions that are defined piecewise, a two-sided limit at a point where theformula for the function changes is best obtained by first finding the one-sided limits atthe point.

Evaluate limx→2

f(x), where f is defined by

f(x) =

x2 − 5x+ 6

|x− 2| for x < 2

2x− 1

x+ 1for x ≥ 2

Example 2.13.

Solution


Suppose that g(x) ≤ f(x) ≤ h(x) for all x in some interval (c, d), except possiblyat the point a ∈ (c, d) and that

limx→a

g(x) = limx→a

h(x) = L

for some number L. Then, it follows that

limx→a

f(x) = L.

0 a x

L

y

g

f

h

Theorem 2.5 (The Squeeze Theorem).

The Squeeze Theorem is also called the Sandwich Theorem or the Pinching Theo-rem.

Evaluate limx→1

f(x) if 8 + 2x− x2 ≤ f(x) ≤ 4x+ 5 for all x ∈ R.

Example 2.14.

Solution

Determine the value of limx→0

[

x cos

(

50π

x

)]

.

Example 2.15.

Solution


Exercise 2.2

1. Given thatlimx→a

f(x) = −3, limx→a

g(x) = 0, limx→a

h(x) = 8

find the limits that exist. If the limit does not exist, explain why.

(a) limx→a

[

f(x) + h(x)]

(b) limx→a

[

f(x)]2

(c) limx→a

3√

h(x) (d) limx→a

1

f(x)

(e) limx→a

f(x)

h(x)(f) lim

x→a

g(x)

f(x)

(g) limx→a

f(x)

g(x)(h) lim

x→a

2f(x)

h(x)− f(x)

2. Evaluate the following limits.

(a) limx→0

(x2 − 3x+ 1) (b) limx→3

(x3 + 2)(x2 − 5x)

(c) limx→2

x− 5

x2 + 4(d) lim

x→1

(

x4 + x2 − 6

x4 + 2x+ 3

)2

(e) limx→1

√x2 + 2x+ 4 (f) lim

x→4−

√16− x2

(g) limx→−3

x2 − x− 12

x+ 3(h) lim

x→−2

x+ 2

x2 − x− 6

(i) limx→1

x2 + x− 2

x2 − 3x+ 2(j) lim

x→1

x3 − 1

x2 − 1

(k) limh→0

(1 + h)4 − 1

h(l) lim

h→0

(2 + h)3 − 8

h

(m) limt→1

t− 1√t− 1

(n) limt→9

9− t

3−√t

(o) limt→2

t2 + t− 6

t2 − 4(p) lim

t→0

√2− t−

√2

t

(q) limx→2

x4 − 16

x− 2(r) lim

x→9

x2 − 81√x− 3

(s) limx→1

[

1

x− 1− 2

x2 − 1

]

(t) limt→0

[

1

t√1 + t

− 1

t

]

(u) limh→0

(3 + h)−1 − 3−1

h(v) lim

x→2

1

x− 1

2

x− 2

(w) limx→1

√x− x

1−√x

(x) limx→0

x√1 + 3x− 1

(y) limx→0

√3 + x−

√3

x(z) lim

x→0

xe−2x+1

x2 + 1

3. Evaluate limx→2

f(x) where f(x) =

{

3x2 − 2x+ 1 if x < 2x3 + 1 if x ≥ 2



f(x) where f(x) =

{

2x if x < 2x2 if x ≥ 2


f(x) where f(x) =

{

x2 + 1 if x < −13x+ 1 if x ≥ −1

6. Evaluate limx→−1

f(x) where f(x) =

2x+ 1 if x < −13 if −1 ≤ x < 12x+ 1 if x ≥ 1

7. Find the limit, if it exists. If the limit does not exist, explain why.

(a) limx→−4

|x+ 4| (b) limx→−4−

|x+ 4|x+ 4

(c) limx→1.5

2x2 − 3x

|2x− 3| (d) limx→0+

(

1

x− 1

|x|

)

8. Use the Squeeze Theorem to find the following limits.

(a) limx→0

x2 sin(1/x)

(b) limx→0

x2 cos(20πx)

(c) limx→0

√x3 + x2 sin

π

x

Answer to Exercise 2.2

1. (a) 5 (b) 9 (c) 2 (d) −1

3(e) −3

8(f) 0 (g) −∞ (h) − 6

11

2. (a) 1 (b) −174 (c) −3

8(d) 4

9(e)

√7 (f) 0 (g) −7 (h) −1

5(i) −3

(j) 3

2(k) 4 (l) 12 (m) 2 (n) 6 (o) 5

4(p) −

√2

4(q) 32 (r) 108 (s) 1

2

(t) −1

2(u) −1

9(v) −1

4(w) 1 (x) 2

3(y) 1

2√3

(z) 0

3. 9 4. 4 5. 1 6. Does not exist 7. (a) 0 (b) −1 (c) Does not exist (d) 0

8. (a) 0 (b) 0 (c) 0


2.3 Limits at Infinity

Let’s begin by investigating the behavior of the function f defined by

f(x) =x2 − 1

x2 + 1

as x becomes large.

x

y

y = 1

x f(x) =x2 − 1

x2 + 1

±0 −1±1 0±2 0.600000±3 0.800000±4 0.882353±5 0.923077±10 0.980198±50 0.999200±100 0.999800±1000 0.999998

As x grows larger and larger we can see from the graph and the table of values thatthe values of f(x) get closer and closer to 1. That is,

limx→∞

x2 − 1

x2 + 1= 1.

In general, we use the notationlimx→∞

f(x) = L

indicates that the values of f(x) becomes closer and closer to L as x becomes larger andlarger.

Let f be a function defined on some interval (a,∞). Then

limx→∞

f(x) = L

mean that the value of f(x) can be made arbitrarily close to L by taking xsufficiently large.

Definition 2.1.

Referring back to the above Figure, we see that for numerically large negative valuesof x, the values of f(x) are close to 1, that is,

limx→−∞

x2 − 1

x2 + 1= 1.

The general definition is as follows.


Let f be a function defined on some interval (−∞, a). Then

limx→−∞

f(x) = L

mean that the value of f(x) can be made arbitrarily close to L by taking xsufficiently large negative.

Definition 2.2.

The line y = L is called a horizontal asymptote of the curve y = f(x) ifeither

limx→∞

f(x) = L or limx→−∞

f(x) = L

Definition 2.3.

For instance, the curve illustrated in Figure above has the line y = 1 as a horizontalasymptote because

limx→∞

x2 − 1

x2 + 1= 1.

For any rational number t > 0,

limx→±∞

1

xt= 0

where for the case where x → −∞, we assume that t =p

qwhere q is odd.

Theorem 2.6.

For a polynomial of degree n > 0,

pn(x) = anxn + an−1x

n−1 + · · ·+ a1x+ a0,

we have

limx→∞

pn(x) =

{

∞, if an > 0−∞, if an < 0

Theorem 2.7.


Evaluate limx→∞

x3 − 2x+ 4

3x2 + 3x− 5.

Example 2.16.

Solution

Evaluate limx→∞

√x4 − 3x2 + 5

2x2 − x.

Example 2.17.

Solution

Evaluate limx→∞

(√x2 + 3x− x

)

.

Example 2.18.

Solution


Evaluate limx→−∞

x3 − 5√x6 − 1− 2x3

.

Example 2.19.

Solution

Exercise 2.3

1− 22 Evaluate the limit, if it exists.

1. limx→∞

x+ 4

x2 − 2x+ 52. lim

x→−∞

(1− x)(2 + x)

(1 + 2x)(2− 3x)

3. limx→−∞

−x√4 + x2

4. limx→−∞

x3 − 2x+ 1

3x3 + 4x− 1

5. limx→∞

x3 − 2x+ 4

3x2 + 3x− 56. lim

x→∞

x2 − sin x

x2 + 4x− 1

7. limx→∞

3x3 − x+ 5

4x3 + 4x2 − 18. lim

x→∞

x4 − x2 + 1

x5 + x3 − x

9. limx→∞

(√x2 + 3− x

)

10. limx→∞

√x2 + 4x

4x+ 1

11. limx→∞

1−√x

1 +√x

12. limx→∞

(√x2 + 1−

√x2 − 1

)

13. limx→∞

(√9x2 + x− 3x

)

14. limx→∞

√x

15. limx→∞

(

x−√x)

16. limx→−∞

(x3 − 5x2)

17. limx→∞

x7 − 1

x6 + 118. lim

x→∞e2x

19. limx→∞

sin 2x 20. limx→∞

e−3x cos 2x

21. limx→∞

ln(2x) 22. limx→0+

(x ln 2x)

23. limx→−∞

x3 − 5√x6 − 1− 2x3

24. limx→−∞

x3 − 1

x3 −√x6 − 1


25− 30 Find the horizontal asymptotes of each curve, if it exists.

25. f(x) =x

x+ 426. f(x) =

x√4 + x2

27. f(x) =x

4− x228. f(x) =

x3

4− x2

29. f(x) =x3

x2 + 3x− 1030. f(x) =

x4√x4 + 1


1. 0 2. 1

63. 1 4. 1

35. ∞ 6. 1 7. 3

48. 0 9. 0 10. 1

411. −1 12. 0

13. 1

614. ∞ 15. ∞ 16. −∞ 17. ∞ 18. ∞ 19. Does not exist 20. 0

21. ∞ 22. 0 23. 1

324. 1

225. y = 1 26. y = −1, y = 1 27. y = 0

28. No horizontal asymptote 29. No horizontal asymptote 30. y = −1, y = 1

2.4 Limit of Trigonometric Functions

limx→0

sin x = 0

Lemma 2.1.

limx→0

sin x

x= 1

Lemma 2.2.

For any real number k 6= 0,

limx→0

sin kx

x= k

Corollary 2.1.

Evaluate limx→0

1− cos 2x

x2.

Example 2.20.

Solution


Evaluate limx→0+

2x− sin x

tan 2x.

Example 2.21.

Solution

Evaluate limx→0

2x cot2 x

csc x.

Example 2.22.

Solution

Evaluate limx→0

1− cosx

x sin x.

Example 2.23.

Solution


Exercise 2.4

1− 10 Evaluate the limits.

1. limx→0

sin 5x

3x2. lim

t→0

sin 8t

sin 9t

3. limθ→0

cos θ − 1

sin θ4. lim

x→0

sin2 x

x

5. limx→0

tan x

4x6. lim

x→0

cot 2x

csc x

7. limx→π/4

sin x− cosx

cos 2x8. lim

x→0

2x cot2 x

csc x

9. limx→0

x+ sin x

tan x10. lim

x→0

sin(cosx)

sec x

11. limx→0

1− cos x

x sin x


1. 5

32. 8

93. 0 4. 0 5. 1

46. 1

27. − 1√

28. 2 9. 2 10. sin 1 11. 1

2

2.5 Continuity

It is helpful for us to first try to see what it is about the function whose graphs areshown below that makes them discontinuous at the point x = a.

x

y

ax

y

b

ax

y

a

This suggests the following definition of continuity at a point.

A function f is continuous at x = a when

(i) f(a) is defined,

(ii) limx→a

f(x) exists and

(iii) limx→a

f(x) = f(a).

Otherwise, f is said to be discontinuous at x = a.

Definition 2.4.


From the graph of f , state the numbers at which f is discontinuous and explainwhy.

x

y

b

0 1 3 5

Example 2.24.

Solution

Where are each of the following functions discontinuous?

(a) f(x) =x5 − 3x2 + 5

x2 − 3x+ 2

(b) g(x) =

2x2 − 5x− 3

x− 3if x 6= 3

6 if x = 3

Example 2.25.

Solution


For what value of a is

f(x) =

{

x2 − 1 if x < 32ax if x ≥ 3

continuous at every x.

Example 2.26.

Solution

Let

f(x) =

(x− 2)2

x2 − 4+ 2k if x > 2

h if x = 2

2x+ k if x < 2

Find the values h and k so that f is continuous at x = 2.

Example 2.27.

Solution


A function f is right-continuous at a if

limx→a+

f(x) = f(a)

and f is left-continuous at a if

limx→a−

f(x) = f(a).

Definition 2.5.

A function f is is said to continuous on a closed interval [a, b] if the followingconditions are satisfied:

1. f is continuous on (a, b).

2. f is right-continuous at a.

3. f is left-continuous at b.

Definition 2.6.

Show that the functionf(x) = x

√16− x2

is continuous on the interval [−4, 4].

Example 2.28.

Solution If −4 < a < 4, then using the Limit Laws, we have

limx→a

f(x) = limx→a

x√16− x2

=(

limx→a

x)(

limx→a

√16− x2

)

=(

limx→a

x)

(

√

limx→a

(16− x2)

)

= a√16− a2 = f(a)

Thus, f is continuous at a if −4 < a < 4. Similar calculations show that

limx→−4+

f(x) = 0 = f(−4) and limx→4−

f(x) = 0 = f(4)

Therefore f is continuous on [−4, 4]. z


If f and g are continuous at x = a and c is a constant, then the followingfunctions are also continuous at x = a:

(a) f + g (b) f − g (c) cf

(d) f · g (e)f

g, provided g(a) 6= 0 (f) fn, n a positive integer

Theorem 2.8.

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

Where is the function h(x) =x4 − 3x+ 1

x2 − x− 6continuous?

Example 2.29.

Solution

If f is continuous at a and g is continuous at f(a), then the composite g ◦ f iscontinuous at a.

Theorem 2.9 (Composite of Continuous Functions).

If g is continuous at L and limx→a

f(x) = L, then

limx→a

g(

f(x))

= g(

limx→a

f(x))

= g(L)

Theorem 2.10 (Limits of Continuous Functions).


Evaluate limx→π/2

cos

(

2x+ sin

(

3π

2+ x

))

.

Example 2.30.

Solution

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

Theorem 2.11 (Intermediate Value Theorem).

f(a)

N

f(b)

y

a c bx

y = f(x)

(a)

f(a)

N

f(b)

y

a c1 c2 c3 bx

y = f(x)

(b)

Note that the value N can be taken on once [as in part (a)] or more that once [as inpart (b)].

Suppose that f is continuous on [a, b] and f(a) and f(b) have opposite signs.Then, there is at least one number c ∈ (a, b) for which f(c) = 0.

Corollary 2.2.


Show that there is a root of the equation x3 − x− 1 = 0 between 1 and 2.

Example 2.31.

Solution Let f(x) = x3 − x − 1. We are looking for a solution of the given equation,that is, a number c between 1 and 2 such that f(c) = 0. Therefore, we take a = 1 andb = 2. We have

f(1) = 13 − 1− 1 = −1 < 0 and f(2) = 23 − 2− 1 = 5 > 0.

Thus f(1) and f(2) have opposite signs. Now f is continuous since it is a polynomial, sothere is a number c ∈ (1, 2) such that f(c) = 0. In other words, the equation x3−x−1 = 0has at least one root c in the interval (1, 2). z

Exercise 2.5

1. Use the given graph to identify all discontinuility of the functions.

(a)

x

y

(b)

x

y

(c)

x

y


2. Use the definition of continuity and the properties of limits to show that the func-tion is continuous at the given number.

(a) f(x) = x2 +√7− x , x = 4

(b) g(x) = (x+ 2x3)4, x = −1

(c) h(x) =x+ 1

2x2 + 1, x = 4

3. Use the definition of continuity and the properties of limits to show that the func-tion is continuous on the given interval.

(a) f(x) =1

x+ 1, (−1,∞)

(b) f(x) =x− 1

x2 − 4, (−2, 2)

(c) g(t) =√9− 4t2 , [−3

2, 3

2]

(d) h(z) =√

(z − 1)(3− z) , [1, 3]

4. Explain why the function is discontinuous at the given number.

(a) f(x) =x

x− 1; x = 1

(b) f(x) = sin1

x; x = 0

(c) f(x) = e1/x ; x = 0

(d) f(x) = ln |x− 2| ; x = 2

(e) f(x) =

{ 1

x− 1if x 6= 1

2 if x = 1; x = 1

(f) f(x) =x2 − 1

x+ 1; x = −1

(g) f(x) =

x2 − 2x− 8

x− 4if x 6= 4

3 if x = 4; x = 4

(h) f(x) =

{

1− x if x ≤ 2x2 − 2x if x > 2

; x = 2

(i) f(x) =

x2 if x < 23 if x = 23x− 2 if x > 2

; x = 2

(j) f(x) =

x2 if x < 0−x if 0 ≤ x ≤ 1x if x > 1

; x = 1


5. Determine the intervals on which f(x) is continuous.

(a) f(x) = 2x+ 3√x (b) f(x) =

1

x+ 3

(c) f(x) =1

x2 + 1(d) f(x) =

x− 5

|x− 5|

(e) f(x) =x2 + 4

x− 2(f) f(x) = 3

√

x+ 1

x− 1

(g) f(x) =3

x2 − x(h) f(x) =

x

4− x2

(i) f(x) =sin x

x2

6. Find the constant c that makes f(x) continuous on (−∞,∞).

(a) f(x) =

{

x+ c if x < 04− x2 if x ≥ 0

(b) f(x) =

{

c2 − x2 if x < 02(x− c)2 if x ≥ 0

(c) f(x) =

{

cx+ 1 if x ≤ 3cx2 − 1 if x > 3

(d) f(x) =

{

x2 − c2 if x < 4cx+ 20 if x ≥ 4

7. For what value of the constant k is the function

f(x) =

x2 − 2x− 3

|x− 3| ; −2 < x < 3

1− kx√x+ 1

; x ≥ 3

continuous on (−2,∞).

8. For what value of the constant k is the function

f(x) =

x3 + 27

x+ 3if x < −3

(x2 − kx)3 if x ≥ −3

continuous on (−∞,∞).

9. Find the values a and b so that f is continuous on (−∞,∞).

f(x) =

x+ 1 if x < 1ax+ b if 1 ≤ x < 23x+ 1 if x ≥ 2


10. Let

f(x) =

{

x2, x 6= 04, x = 0

and g(x) = 2x.

Show thatlimx→0

f(

g(x))

6= f(

limx→0

g(x))

.

11. Use the Intermediate Value Theorem to verify that f(x) has a zero in the giveninterval.

(a) f(x) = x2 − 7 ; [2, 3]

(b) f(x) = x3 − 4x− 2 ; [−1, 0]

(c) f(x) = cosx− x ; [0, 1]

12. Use the Intermediate Value Theorem to show that x3 + 3x− 2 = 0 has a real rootbetween 0 and 1.

13. Use the Intermediate Value Theorem to show that (cos t) t3 + 6 sin5 t − 3 = 0 hasa real root between 0 and 2π.

14. Show that the equation x5 + 4x3 − 7x+ 14 = 0 has at least one real root.


1. (a) x = −2, 2 (b) x = −2, 1, 4 (c) x = −2, 2, 4

4. (a) f(1) is not defined (b) f(0) is not defined (c) f(0) is not defined

(d) f(2) is not defined (e) limx→1

f(x) is not defined (f) f(−1) is not defined

(g) limx→4

f(x) 6= f(4) (h) limx→2

f(x) is not defined (i) limx→2

f(x) 6= f(2)

(j) limx→1

f(x) is not defined

5. (a) R (b) (−∞,−3)∪(−3,∞) (c) R (d) (−∞, 5)∪(5,∞) (e) (−∞, 2)∪(2,∞)

(f) (−∞, 1)∪(1,∞) (g) (−∞, 0)∪(0, 1)∪(1,∞) (h) (−∞,−2)∪(−2, 2)∪(2,∞)

(i) (−∞, 0) ∪ (0,∞)

6. (a) 4 (b) 0 (c) 1

3(d) −2 7. 3 8. −2 9. a = 5, b = −3

chapter 2 limits and continuitymathstat.sci.tu.ac.th/~archara/ma111/ma111-118/... · limits and...

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