Adams Calculus 8th Edition Solution – Test Bank

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INSTRUCTOR’S SOLUTIONS MANUAL SECTION 1.1 (PAGE 63)

CHAPTER 1. LIMITS AND CONTINUITY

Section 1.1 Examples of Velocity, Growth

Rate, and Area (page 63)

1. Average velocity =

x

t

=

(t + h)2

h

− t2

m/s.

7. 8. At t = 1 the velocity is v = −6 < 0 so the particle is

moving to the left.

At t = 2 the velocity is v = 0 so the particle is station-

ary.

At t = 3 the velocity is v = 6 > 0 so the particle is

moving to the right.

Average velocity over [t− k, t + k] is

3(t + k)2

− 12(t + k) + 1− [3(t− k)2

− 12(t− k) + 1]

2.

3. h Avg. vel. over [2, 2 + h]

1 5.0000

0.1 4.1000

0.01 4.0100

0.001 4.0010

0.0001 4.0001

Guess velocity is v = 4 m/s at t = 2 s.

4. Average volocity on [2, 2 + h] is

(2 + h)2

− 4

(2 + h)− 2=

4 + 4h + h2

− 4

h=

4h + h2

h= 4 + h.

As h approaches 0 this average velocity approaches 4

m/s

5. x = 3t2

− 12t + 1 m at time t s.

Average velocity over interval [1, 2] is

(3 × 22

− 12 × 2 + 1)− (3 × 12

− 12 × 1 + 1)

2− 1= −3

m/s.

Average velocity over interval [2, 3] is

(3 × 32

− 12 × 3 + 1)− (3 × 22

− 12 × 2 + 1)

3− 2= 3 m/s.

Average velocity over interval [1, 3] is

(3 × 32

− 12 × 3 + 1)− (3 × 12

− 12 × 1 + 1)

3− 1= 0 m/s.

6. Average velocity over [t, t + h] is

3(t + h)2

− 12(t + h) + 1− (3t2

− 12t + 1)

9.

(t + k)− (t− k)

=

1

2k

3t2 + 6tk + 3k2

− 12t− 12k + 1− 3t2 + 6tk− 3k2

=

+ 12t− 12k + 1

12tk− 24k

2k= 6t− 12 m/s,

which is the velocity at time t from Exercise 7.

y

2

1

1

y= 2 +

sin(π t)

π

10. 1 2 3 4 5

Fig. 1.1.9

t

At t = 1 the height is y= 2 ft and the weight is

moving downward.

Average velocity over [1, 1 + h] is

1

2 +

π

sin π(1 + h)− 2 +

1

π

sin π

=

h

= −

sin(π + π h)

π h=

sin(π h)

π h.

sin π cos(π h) + cos π sin(π h)

π h

=

(t + h)− t

6th + 3h2

− 12h

h= 6t + 3h− 12 m/s.

This average velocity approaches 6t− 12 m/s as h ap-

proaches 0.

At t = 1 the velocity is 6 × 1− 12= −6 m/s.

At t = 2 the velocity is 6 × 2− 12= 0 m/s.

At t = 3 the velocity is 6 × 3− 12= 6 m/s.

h Avg. vel. on [1, 1 + h]

1.0000 0

0.1000 -0.983631643

0.0100 -0.999835515

0.0010 -0.999998355

11. The velocity at t = 1 is about v = −1 ft/s. The “−

”

indicates that the weight is moving downward.

23

Copyright © 2014 Pearson Canada Inc.SECTION 1.1 (PAGE 63) ADAMS and ESSEX: CALCULUS 8

12. We sketched a tangent line to the graph on page 55 in

the text at t = 20. The line appeared to pass through

the points (10, 0) and (50, 1). On day 20 the biomass is

growing at about (1− 0)/(50− 10)= 0.025 mm2/d.

13. The curve is steepest, and therefore the biomass is grow-

ing most rapidly, at about day 45.

14. a) profit

175

150

125

100

75

50

25

y

1

y= g(x)

x

1 2 3

year

2008 2009 2010 2011 2012

Fig. 1.1.14

b) Average rate of increase in profits between 2010 and

2012 is

174− 62

2012− 2010=

112

2= 56 (thousand$/yr).

c) Drawing a tangent line to the graph in (a) at

t = 2010 and measuring its slope, we find that

the rate of increase of profits in 2010 is about 43

thousand$/year.

Section 1.2 Limits of Functions (page 71)

1. From inspecting the graph

y

y= f (x)

1

x

−1 1

Fig. 1.2.1

we see that

lim

x→−1 f (x)= 1, lim

x→0 f (x)= 0, lim

x→1 f (x)= 1.

Fig. 1.2.2

we see that

lim

x→1 g(x) does not exist

(left limit is 1, right limit is 0)

lim

x→2 g(x)= 1, lim

x→3 g(x)= 0.

3. lim

x→1−

g(x)= 1

4. lim

x→1+

g(x)= 0

5. lim

x→3+

g(x)= 0

6. lim

x→3−

g(x)= 0

7. lim

x→4(x2

− 4x + 1)= 42

− 4(4) + 1= 1

8. lim

x→2

3(1− x)(2− x)= 3(−1)(2− 2)= 0

9. lim

x→3

x + 3

x + 6=

3 + 3

3 + 6=

2

3

t2

10. lim

t→−4

=

4− t

(−4)2

4 + 4= 2

11. lim

x→1

x2

− 1

x + 1=

12

− 1

1 + 1=

0

2= 0

12. lim

x→−1

x2

− 1

x + 1= lim

x→−1(x− 1)= −2

x2

13. lim

x→3

= lim

x→3

− 6x + 9

x2

− 9= lim

x→3

x− 3

0

x + 3=

6= 0

(x− 3)2

(x− 3)(x + 3)

14. lim

x→−2

x2 + 2x

x2

− 4= lim

x→−2

x

x− 2=

−2

−4=

1

2

1

15. limh→2

4− h2 does not exist; denominator approaches 0

but numerator does not approach 0.

16. limh→0

3h + 4h2

h2

− h3

3 + 4h

= lim

h→0

h− h2 does not exist; denomi-

nator approaches 0 but numerator does not approach 0.

2. From inspecting the graph

24

Copyright © 2014 Pearson Canada Inc.INSTRUCTOR’S SOLUTIONS MANUAL SECTION 1.2 (PAGE 71)

17. lim

x→9

√x− 3

x− 9= lim

x→9

= lim

x→9

(√x− 3)(√x + 3)

(x− 9)(√x + 3)

x− 9

(x− 9)(√x + 3)

= lim

x→9

√4 + h− 2

18. lim

h→0

h

4 + h− 4

= lim

h→0

h(√4 + h + 2)

1

1

= lim

h→0

√4 + h + 2=

4

(x− π )2

02

19. lim

=

= 0

x→π

π x

π 2

20. lim

x→−2 |x− 2| = | − 4| = 4

21. lim

x→0

|x− 2|

x− 2=

| − 2|

−2= −1

22. lim

x→2

|x− 2|

x− 2= lim

x→2

|x− 2|

x− 2

1, if x > 2

−1, if x < 2.

Hence, lim

x→2

t2

− 1

does not exist.

23. lim

t→1

t2

− 2t + 1

(t− 1)(t + 1)

t + 1

lim

t→1

= lim

(t− 1)2

t→1

t− 1

(denominator → 0, numerator → 2.)

does not exist

24. lim

x→2

= lim

x→2

25. lim

t→0

√4− 4x + x2

x− 2

|x− 2|

x− 2

t

√4 + t− √4− t

does not exist.

= lim

t→0

= lim

t→0

t(√4 + t + √4− t)

(4 + t)− (4− t)

√4 + t + √4− t

2= 2

26. lim

x→1

x2

− 1

(x− 1)(x + 1)(√x + 3 + 2)

√x + 3− 2= lim

x→1

(x + 3)− 4

= lim

x→1(x + 1)(√x + 3 + 2)= (2)(√4 + 2)= 8

27. lim

t→0

= lim

t→0

= lim

t→0

t2 + 3t

(t + 2)2

− (t− 2)2

t(t + 3)

t2 + 4t + 4− (t2

t + 3

3

8=

8

− 4t + 4)

(s + 1)2

− (s− 1)2

4s

28. lim

s→0

= lim

s→0

29. lim

y→1

= lim

y→1

s

s

y− 4√y + 3

y2

− 1

(√y− 1)(√y− 3)

(√y− 1)(√y + 1)(y + 1)

1

√x + 3=

1

6

= 4

=

−2

4=

−1

2

30. lim

x→−1

= lim

x→−1

x3 + 1

x + 1

(x + 1)(x2

− x + 1)

x + 1= 3

31. lim

x→2

x4

x3

− 16

− 8

(x− 2)(x + 2)(x2 + 4)

= lim

x→2

(x− 2)(x2 + 2x + 4)

(4)(8)

8

=

4 + 4 + 4=

3

32. lim

x→8

x2/3

− 4

x1/3

− 2

(x1/3

− 2)(x1/3 + 2)

= lim

x→8

(x1/3

− 2)

= lim

x→8(x1/3 + 2)= 4

33. lim

x→2

1

4

x− 2−

x2

− 4

x + 2− 4

= lim

x→2

(x− 2)(x + 2)

= lim

x→2

34. lim

x→2

1

1

x− 2−

x2

− 4

x + 2− 1

= lim

x→2

(x− 2)(x + 2)

x + 1

= lim

x→2

(x− 2)(x + 2) does not exist.

√2 + x2

− √2− x2

35. lim

x→0

x2

(2 + x2)− (2− x2)

= lim

x→0

x2(√2 + x2 + √2− x2)

2x2

= lim

x→0

x2 √2 + x2) + √2− x2

=

2

√2 + √2=

1

√2

|3x− 1| − |3x + 1|

36. lim

x→0

= lim

x→0

x

(3x− 1)2

− (3x + 1)2

x (|3x− 1| + |3x + 1|)

−12x

= lim

x→0

x (|3x− 1| + |3x + 1|)

37. f (x)= x2

f (x + h)− f (x)

lim

h→0

h= lim

h→0

= lim

h→0

1

x + 2=

1

4

=

−12

1 + 1= −6

(x + h)2

− x2

h

2hx + h2

h= lim

h→0

2x + h= 2x

25

Copyright © 2014 Pearson Canada Inc.ADAMS and ESSEX: CALCULUS 8

(x + h)3

− x3

SECTION 1.2 (PAGE 71) 38. f (x)= x3

f (x + h)− f (x)

lim

h→0

h= lim

h→0

= lim

h→0

= lim

h→0

h

3x2 + 3xh + h2

39. f (x)= 1/x

1

1

lim

h→0

f (x + h)− f (x)

h= lim

h→0

= lim

h→0

x + h−

x

h

x− (x + h)

h(x + h)x

1

= lim

h→0

−

= −

(x + h)x

40. f (x)= 1/x2

1

(x + h)2−

1

x2

lim

h→0

f (x + h)− f (x)

h= lim

h→0

= lim

h→0

= lim

h→0

46. lim

x→2π/3

sin x = sin 2π/3= √3/2

47.

h

3x2h + 3xh2 + h3

x (sin x)/x

= 3x2

1

±1.0 0.84147098

±0.1 0.99833417

±0.01 0.99998333

±0.001 0.99999983

0.0001 1.00000000

sin x

It appears that lim

x→0

x

= 1.

48.

x (1− cos x)/x2

x2

h

x2

− (x2 + 2xh + h2)

±1.0 0.45969769

±0.1 0.49958347

±0.01 0.49999583

±0.001 0.49999996

0.0001 0.50000000

1

=

−

h(x + h)2x2

2x + h

= −

(x + h)2x2

2x

x4

√x + h− √x

1

2

= −

x3

2√x

41. f (x)= √x

f (x + h)− f (x)

lim

h→0

h= lim

h→0

= lim

h→0

h

x + h− x

h(√x + h + √x)

1

= lim

h→0

=

√x + h + √x

42. f (x)= 1/√x

1

1

−

lim

h→0

f (x + h)− f (x)

h= lim

h→0

√x + h

√x

= lim

h→0

= lim

h→0

h

√x− √x + h

h√x√x + h

x− (x + h)

h√x√x + h(√x + √x + h)

−1

= lim

h→0

√x√x + h(√x + √x + h)

=

−1

2x3/2

sin x = sin π/2= 1

cos x = cos π/4= 1/√2

cos x = cos π/3= 1/2

43. lim

x→π/2

44. lim

x→π/4

45. lim

x→π/3

26

1− cos x

It appears that lim

x→0

x2

.

2

49. lim

x→2−

√2− x = 0

50. lim

x→2+

√2− x does not exist.

51. lim

x→−2−

√2− x = 2

52. lim

x→−2+

√2− x = 2

53. lim

x→0

x3

(x3

− x does not exist.

− x < 0 if 0 < x < 1)

54. lim

x→0−

x3

− x = 0

55. lim

x→0+

x3

− x does not exist. (See # 9.)

56. lim

x→0+

x2

− x4 = 0

57. lim

x→a−

|x− a|

x2

− a2

= lim

x→a−

|x− a|

(x− a)(x + a)

= −

2a

58. lim

x→a+

|x− a|

x2

− a2

= lim

x→a+

x− a

x2

− a2

1

x2

− 4

0

59. lim

x→2−

=

|x + 2|

4= 0

x2

− 4

0

60. lim

x→2+

=

|x + 2|

4= 0

1

(a ̸= 0)

=

2a

Copyright © 2014 Pearson Canada Inc.INSTRUCTOR’S SOLUTIONS MANUAL 61. f (x)=

x− 1 if x ≤ −1

x2 + 1 if−1 < x ≤ 0

(x + π )2 if x > 0

lim

f (x)= lim

x− 1= −1− 1= −2

x→−1−

x→−1−

62. lim

f (x)= lim

x2 + 1= 1 + 1= 2

x→−1+

x→−1+

63. lim

x→0+

f (x)= lim

x→0+

(x + π )2

= π 2

64. lim

x→0−

f (x)= lim

x→0−

x2 + 1= 1

65. If lim

x→4 f (x)= 2 and lim

x→4 g(x)= −3, then

a) lim

x→4 g(x) + 3= −3 + 3= 0

b) lim

x→4 x f (x)= 4 × 2= 8

c) lim

x→4 g(x)

2

= (−3)2

= 9

d) lim

x→4

g(x)

f (x)− 1=

−3

2− 1= −3

66. If lim x → a f (x)= 4 and lim

g(x)= −2, then

x→a

a) lim

x→a

f (x) + g(x)= 4 + (−2)= 2

b) lim

x→a

f (x)· g(x)= 4 × (−2)= −8

c) lim

x→a

4g(x)= 4(−2)= −8

d) lim

x→a

f (x)

g(x)

4

=

−2= −2

67. If lim

x→2

f (x)− 5

x− 2= 3, then

lim

x→2 f (x)− 5= lim

x→2

f (x)− 5

x− 2 (x− 2)= 3(2− 2)= 0.

Thus limx→2 f (x)= 5.

f (x)

68. If lim

= −2 then

x→0

x2

limx→0 f (x)= limx→0 x2 f (x)

x2

and similarly,

f (x)

limx→0

x

= lim

x→0

x

f (x)

x2

= 0 × (−2)= 0,

= 0 × (−2)= 0.

SECTION 1.2 (PAGE 71)

69.

y

1.2

1.0

0.8

0.6

0.4

0.2

sin x

y=

x

-3 -2 -1 1 2

x

-0.2

-0.4

lim

x→0

sin x

x

= 1

70.

Fig. 1.2.69

y

0.8

0.6

0.4

0.2

sin(2π x)

y=

sin(3π x)

-0.08 -0.04 0.04 0.08

x

-0.2

-0.4

71.

Fig. 1.2.70

limx→0 sin(2π x)/ sin(3π x)= 2/3

y

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

sin √1− x

y=

√1− x2

0.2 0.4 0.6 0.8 1.0

x

-0.1

Fig. 1.2.71

lim

x→1−

sin √1− x

√1− x2

≈ 0.7071

27

Copyright © 2014 Pearson Canada Inc.72.

73.

74. 75. SECTION 1.2 (PAGE 71) y

76. a)

0.2 0.4 0.6 0.8

y=

x− √x

√sin x

x

(−1, 1) ADAMS and ESSEX: CALCULUS 8

y

3

2

1

-0.2

-0.4

-0.6

-0.8

-1.0

-1.2

lim

x→0+

y= x4

(1, 1)

y= x2

x

x− √x

√sin x

Fig. 1.2.72

= −1

y

y= −x

y= x sin(1/x)

0.1

y= x

-0.2 -0.1 0.1

x

-0.1

-0.2

Fig. 1.2.73

f (x)= x sin(1/x) oscillates infinitely often as x ap-

proaches 0, but the amplitude of the oscillations decreases

and, in fact, limx→0 f (x)= 0. This is predictable be-

cause |x sin(1/x)| ≤ |x|. (See Exercise 95 below.)

Since √5− 2x2 ≤ f (x) ≤ √5− x2 for−1 ≤ x ≤ 1, and

limx→0 √5− 2x2 = limx→0 √5− x2 = √5, we have

limx→0 f (x)= √5 by the squeeze theorem.

Since 2− x2 ≤ g(x) ≤ 2 cos x for all x, and since

limx→0(2− x2)= limx→0 2 cos x = 2, we have

limx→0 g(x)= 2 by the squeeze theorem.

28

-2 -1 1

Fig. 1.2.76

b) Since the graph of f lies between those of x2 and

x4, and since these latter graphs come together at

(±1, 1) and at (0, 0), we have limx→±1 f (x)= 1

and limx→0 f (x)= 0 by the squeeze theorem.

77. x1/3 < x3 on (−1, 0) and (1, ∞). x1/3 > x3 on

(−∞,−1) and (0, 1). The graphs of x1/3 and x3 inter-

sect at (−1,−1), (0, 0), and (1, 1). If the graph of h(x)

lies between those of x1/3 and x3, then we can determine

limx→a h(x) for a = −1, a = 0, and a = 1 by the

squeeze theorem. In fact

lim

x→−1

h(x)= −1, lim

x→0

h(x)= 0, lim

x→1

h(x)= 1.

1

78. f (x)= s sin

is defined for all x ̸= 0; its domain is

x

(−∞, 0) ∪ (0, ∞). Since | sin t| ≤ 1 for all t, we have

| f (x)| ≤ |x| and −|x| ≤ f (x) ≤ |x| for all x ̸= 0.

Since limx→0 = (−|x|)= 0= limx→0 |x|, we have

limx→0 f (x)= 0 by the squeeze theorem.

79. | f (x)| ≤ g(x) ⇒ −g(x) ≤ f (x) ≤ g(x)

Since lim

g(x)= 0, therefore 0 ≤ lim

x→a

x→a

Hence, lim

f (x)= 0.

x→a

If lim

g(x)= 3, then either−3 ≤ lim

x→a

x→a

limx→a f (x) does not exist.

f (x) ≤ 0.

f (x) ≤ 3 or

Section 1.3 Limits (page 78)

Limits at Infinity and Infinite

1. lim

x→∞

x

2x− 3= lim

x→∞

1

2− (3/x)

=

1

2

2. lim

x→∞

x2

x

− 4= lim

x→∞

1/x

1− (4/x2)

0

=

1= 0

Copyright © 2014 Pearson Canada Inc.SECTION 1.3 (PAGE 78)

1

3

= −

5

INSTRUCTOR’S SOLUTIONS MANUAL 3x3

− 5x2 + 7

3. lim

x→∞

8 + 2x− 5x3

5

7

3−

+

= lim

x→∞

x

x3

8

2

x3 +

x2− 5

x2

− 2

4. lim

x→−∞

x− x2

1−

2

x2

= lim

x→−∞

=

1

1

−1= −1

x− 1

x2 + 3

5. lim

x→−∞

x3 + 2= lim

x→−∞

1

+

x

1 +

6. lim

x→∞

x2 + sin x

x2 + cos x

1 +

= lim

x→∞

1 +

3

x3

2

x3

sin x

x2

cos x

x2

= 0

We have used the fact that limx→∞

1

=

1= 1

sin x

x2

= 0 (and simi-

larly for cosine) because the numerator is bounded while

the denominator grows large.

7. lim

x→∞

3x + 2√x

1− x

= lim

x→∞

2

3 +

1

√x

x− 1

= −3

8. lim

x→∞

= lim

x→∞

2x− 1

√3x2 + x + 1

x 2−

|x| 3 +

1

x

1

1

x

+

= lim

x→∞

2−

1

x

3 +

+

x

1

x2

2x− 1

9. lim

x→−∞

√3x2 + x + 1

2−

1

x

= lim

= −

x→−∞

− 3 +

1

x

√3

,

+

x2

because x → −∞ implies that x < 0 and so √x2 = −x.

2x− 5

2x− 5

10. lim

x→−∞

|3x + 2|

= lim

x→−∞

= −

−(3x + 2)

2

3

1

11. lim

x→3

3− x

does not exist.

(but |x| = x as x → ∞)

1

x2

=

2

√3

2

1

12. lim

x→3

= ∞

13. lim

x→3−

14. lim

x→3+

(3− x)2

1

3− x

1

3− x

= ∞

= −∞

15. lim

x→−5/2

2x + 5

5x + 2=

0

= 0

−25

2 + 2

16. lim

x→−2/5

2x + 5

5x + 2

does not exist.

2x + 5

17. lim

x→−(2/5)−

5x + 2

= −∞

18. lim

x→−2/5+

2x + 5

5x + 2

= ∞

19. lim

x→2+

20. lim

x→1−

21. lim

x→1+

22. lim

x→1−

x

(2− x)3

x

√1− x2

1

|x− 1|

1

|x− 1|

= −∞

= ∞

= ∞

= ∞

23. lim

x→2

x2

x− 3

− 4x + 4= lim

x→2

24. lim

x→1+

√x2

− x

x− x2

= lim

x→1+

√x2

− x

25. lim

x→∞

= lim

x→∞

x + x3 + x5

1 + x2 + x3

1

x2 + 1 + x2

1

1

x3 +

+ 1

x

= ∞

26. lim

x→∞

x3 + 3

x2 + 2= lim

x→∞

x +

3

x2

27. lim

x→∞

= lim

x→∞

x√x + 1 1− √2x + 3

7− 6x + 4x2

x2 1 +

1

x

x2 7

x2−

1

4 √2

1

√x− 2 +

6

x

+ 4

1(−√2)

=

4= −

28. lim

x→∞

x2

x + 1−

x2

x− 1= lim

x→∞

x− 3

(x− 2)2 = −∞

−1

= −∞

= ∞

1 +

2

x2

3

x

−2x2

x2

− 1= −2

29

Copyright © 2014 Pearson Canada Inc.SECTION 1.3 (PAGE 78) ADAMS and ESSEX: CALCULUS 8

2

x

2

x

=

4

2= 2

29. lim

x→−∞

x2 + 2x− x2

− 2x

− 2x)

= lim

x→−∞

(x2 + 2x)− (x2

√x2 + 2x + √x2

4x

− 2x

= lim

x→−∞

(−x) 1 +

2

x

+ 1−

4

= −

1 + 1= −2

30. lim

x→∞

x2 + 2x− x2

− 2x

x2 + 2x− x2 + 2x

= lim

x→∞

√x2 + 2x + √x2

− 2x

4x

= lim

x→∞

2

= lim

x→∞

31. lim

x→∞

√x2

x 1 +

x

4

2

1 +

x

+ 1−

1

− 2x− x

+ x 1−

2

x

= lim

x→∞

= lim

x→∞

= lim

x→∞

1

32. lim

x→−∞

√x2 + 2x− x

√x2

− 2x + x

(√x2

√x2

x2

− 2x + x)(√x2

− 2x− x)

− 2x + x

− 2x− x2

x(√1− (2/x) + 1)

=

−2x

= lim

x→−∞

2

−2= −1

1

|x|(√1 + (2/x) + 1= 0

33. By Exercise 35, y= −1 is a horizontal asymptote (at the

1

right) of y=

√x2

− 2x− x

. Since

1

lim

x→−∞

√x2

− 2x− x

= lim

x→−∞

1

|x|(√1− (2/x) + 1= 0,

y= 0 is also a horizontal asymptote (at the left).

Now √x2

− 2x− x = 0 if and only if x2

− 2x = x2, that

is, if and only if x = 0. The given function is undefined

at x = 0, and where x2

− 2x < 0, that is, on the interval

[0, 2]. Its only vertical asymptote is at x = 0, where

1

limx→0−

= ∞.

√x2

− 2x− x

2x− 5

2

2x− 5

34. Since lim

=

and lim

= −

x→∞

|3x + 2|

3

x→−∞

|3x + 2|

y = ±(2/3) are horizontal asymptotes of

y= (2x− 5)/|3x + 2|. The only vertical asymptote

is x = −2/3, which makes the denominator zero.

2

3 ,

35. lim

x→0+

f (x)= 1

30

36. lim

x→1 f (x) = ∞

37.

y

3

2

1

y= f (x)

1 2 3 4 5 6

x

-1

Fig. 1.3.37

limx→2+ f (x)= 1

38. lim

x→2−

f (x)= 2

39. lim

x→3−

f (x) = −∞

40. lim

x→3+

f (x) = ∞

41. lim

x→4+

f (x)= 2

42. lim

x→4−

f (x)= 0

43. lim

x→5−

f (x)= −1

44. lim

x→5+

f (x)= 0

45. lim

x→∞

f (x)= 1

46. horizontal: y= 1; vertical: x = 1, x = 3.

47. lim

x→3+

⌊x⌋ = 3

48. lim

x→3−

⌊x⌋ = 2

49. lim

x→3⌊x⌋ does not exist

50. lim

x→2.5

⌊x⌋ = 2

51. lim

x→0+

⌊2− x⌋ = lim

x→2−

⌊x⌋ = 1

52. lim

x→−3−

⌊x⌋ = −4

53. lim

t→t0

lim

t→t0−

lim

t→t0+

lim

t→t0+

C(t)= C(t0) except at integers t0

C(t)= C(t0) everywhere

C(t)= C(t0) if t0 ̸= an integer

C(t)= C(t0) + 1.5 if t0 is an integer

Copyright © 2014 Pearson Canada Inc.INSTRUCTOR’S SOLUTIONS MANUAL y

6.00

4.50

3.00

1.50

SECTION 1.4 (PAGE 87)

y= C(t)

3. g has no absolute maximum value on [−2, 2]. It takes

on every positive real value less than 2, but does not take

the value 2. It has absolute minimum value 0 on that

interval, assuming this value at the three points x = −2,

x = −1, and x = 1.

4. Function f is discontinuous at x = 1, 2, 3, 4, and 5. f

is left continuous at x = 4 and right continuous at x = 2

and x = 5.

1 2 3 4

x

Fig. 1.3.53

54. lim

f (x)= L

x→0+

(a) If f is even, then f (−x)= f (x).

Hence, lim

f (x)= L.

x→0−

(b) If f is odd, then f (−x)= − f (x).

Therefore, lim

f (x)= −L.

x→0−

55. lim

x→0+

f (x)= A, lim

x→0−

f (x)= B

a) lim

x→0+

f (x3

− x)= B (since x3

− x < 0 if 0 < x < 1)

b) lim

f (x3

x→0−

−1 < x < 0)

c) lim

x→0−

f (x2

− x)= A (because x3

− x > 0 if

− x4)= A

d) lim

f (x2

x→0+

0 < |x| < 1)

− x4)= A (since x2

− x4 > 0 for

Section 1.4 Continuity (page 87)

1. g is continuous at x = −2, discontinuous at

x = −1, 0, 1, and 2. It is left continuous at x = 0

and right continuous at x = 1.

y

3

2

1

y= f (x)

1 2 3 4 5 6

x

-1

(−1, 1)

y

2

1

(1, 2)

y= g(x)

-2 -1 1 2

x

Fig. 1.4.1

2. g has removable discontinuities at x = −1 and x = 2.

Redefine g(−1)= 1 and g(2)= 0 to make g continuous

at those points.

Fig. 1.4.4

5. f cannot be redefined at x = 1 to become continuous

there because limx→1 f (x) (= ∞) does not exist. (∞ is

not a real number.)

6. sgn x is not defined at x = 0, so cannot be either continu-

ous or discontinuous there. (Functions can be continuous

or discontinuous only at points in their domains!)

7. f (x)=

x if x < 0

x2 if x ≥ 0 is continuous everywhere on the

real line, even at x = 0 where its left and right limits are

both 0, which is f (0).

8. f (x)=

x if x <−1

x2 if x ≥ −1 is continuous everywhere on the

real line except at x = −1 where it is right continuous,

but not left continuous.

lim

x→−1−

f (x)= lim

x = −1 ̸= 1

x→−1−

= f (−1)= lim

x2

x→−1+

= lim

x→−1+

f (x).

9. f (x)=

1/x2 if x ̸= 0

0 if x = 0 is continuous everywhere ex-

cept at x = 0, where it is neither left nor right continuous

since it does not have a real limit there.

10. f (x)=

x2 if x ≤ 1

is continuous everywhere

0.987 if x > 1

except at x = 1, where it is left continuous but not right

continuous because 0.987 ̸= 1. Close, as they say, but no

cigar.

31

Copyright © 2014 Pearson Canada Inc.SECTION 1.4 (PAGE 87) ADAMS and ESSEX: CALCULUS 8

11. The least integer function ⌈x⌉ is continuous everywhere

on except at the integers, where it is left continuous

but not right continuous.

12. 13. Since

14. Since

C(t) is discontinuous only at the integers. It is continu-

ous on the left at the integers, but not on the right.

x2

− 4

x− 2= x + 2 for x ̸= 2, we can define the

function to be 2 + 2= 4 at x = 2 to make it continuous

there. The continuous extension is x + 2.

1 + t3

(1 + t)(1− t + t2)

1− t + t2

=

=

for

1− t2

(1 + t)(1− t)

1− t

t ̸= −1, we can define the function to be 3/2 at t = −1

to make it continuous there. The continuous extension is

1− t + t2

1− t.

15. Since

t2

− 5t + 6

(t− 2)(t− 3)

t− 2

t2

− t− 6=

=

for t ̸= 3,

(t + 2)(t− 3)

t + 2

we can define the function to be 1/5 at t = 3 to make it

t− 2

continuous there. The continuous extension is

.

t + 2

16. Since

x2

x4

− 2

(x− √2)(x + √2)

x + √2

− 4=

=

(x− √2)(x + √2)(x2 + 2)

(x + √2)(x2 + 2)

for x ̸= √2, we can define the function to be 1/4 at

x = √2 to make it continuous there. The continuous

x + √2

extension is

(x + √2)(x2 + 2) . (Note: cancelling the

x + √2 factors provides a further continuous extension to

x = −√2.

17. limx→2+ f (x)= k− 4 and limx→2− f (x)= 4= f (2).

Thus f will be continuous at x = 2 if k− 4= 4, that is,

if k= 8.

18. limx→3− g(x)= 3− m and

limx→3+ g(x)= 1− 3m = g(3). Thus g will be con-

tinuous at x = 3 if 3− m = 1− 3m, that is, if m = −1.

19. x2 has no maximum value on−1 < x < 1; it takes all

positive real values less than 1, but it does not take the

value 1. It does have a minimum value, namely 0 taken

on at x = 0.

20. The Max-Min Theorem says that a continuous function

defined on a closed, finite interval must have maximum

and minimum values. It does not say that other functions

cannot have such values. The Heaviside function is not

continuous on [−1, 1] (because it is discontinuous at

x = 0), but it still has maximum and minimum values.

Do not confuse a theorem with its converse.

21. Let the numbers be x and y, where x ≥ 0, y ≥ 0, and

x + y= 8. If P is the product of the numbers, then

P= xy= x(8− x)= 8x− x2

= 16− (x− 4)2

.

Therefore P ≤ 16, so P is bounded. Clearly P= 16 if

x = y= 4, so the largest value of P is 16.

22. Let the numbers be x and y, where x ≥ 0, y ≥ 0, and

x + y= 8. If S is the sum of their squares then

S= x2 + y2

= x2 + (8− x)2

= 2x2

− 16x + 64= 2(x− 4)2 + 32.

Since 0 ≤ x ≤ 8, the maximum value of S occurs at

x = 0 or x = 8, and is 64. The minimum value occurs at

x = 4 and is 32.

23. Since T= 100− 30x + 3x2

= 3(x− 5)2 + 25, T will

be minimum when x = 5. Five programmers should be

assigned, and the project will be completed in 25 days.

24. If x desks are shipped, the shipping cost per desk is

245x− 30x2 + x3

C=

= x2

− 30x + 245

x

= (x− 15)2 + 20.

This cost is minimized if x = 15. The manufacturer

should send 15 desks in each shipment, and the shipping

cost will then be $20 per desk.

x2

− 1

(x− 1)(x + 1)

25. f (x)=

=

x

x

f= 0 at x = ±1. f is not defined at 0.

f (x) > 0 on (−1, 0) and (1, ∞).

f (x) < 0 on (−∞,−1) and (0, 1).

26. f (x)= x2 + 4x + 3= (x + 1)(x + 3)

f (x) > 0 on (−∞,−3) and (−1, ∞)

f (x) < 0 on (−3,−1).

x2

− 1

27. f (x)=

x2

− 4=

(x− 1)(x + 1)

(x− 2)(x + 2)

f= 0 at x = ±1.

f is not defined at x = ±2.

f (x) > 0 on (−∞,−2), (−1, 1), and (2, ∞).

f (x) < 0 on (−2,−1) and (1, 2).

x2 + x− 2

(x + 2)(x− 1)

28. f (x)=

=

x3

x3

f (x) > 0 on (−2, 0) and (1, ∞)

f (x) < 0 on (−∞,−2) and (0, 1).

29. f (x)= x3 + x− 1, f (0)= −1, f (1)= 1.

Since f is continuous and changes sign between 0 and 1,

it must be zero at some point between 0 and 1 by IVT.

30. f (x)= x3

− 15x + 1 is continuous everywhere.

f (−4)= −3, f (−3)= 19, f (1)= −13, f (4)= 5.

Because of the sign changes f has a zero between−4

and−3, another zero between−3 and 1, and another

between 1 and 4.

32

Copyright © 2014 Pearson Canada Inc.INSTRUCTOR’S SOLUTIONS MANUAL SECTION 1.5 (PAGE 92)

31. F(x)= (x− a)2(x− b)2 + x. Without loss of generality,

we can assume that a < b. Being a polynomial, F is

continuous on [a, b]. Also F(a)= a and F(b)= b.

Since a < 1

2 (a + b) < b, the Intermediate-Value Theorem

guarantees that there is an x in (a, b) such that

F(x)= (a + b)/2.

32. Let g(x)= f (x)− x. Since 0 ≤ f (x) ≤ 1 if 0 ≤ x ≤ 1,

therefore, g(0) ≥ 0 and g(1) ≤ 0. If g(0)= 0 let c = 0,

or if g(1)= 0 let c = 1. (In either case f (c)= c.)

Otherwise, g(0) > 0 and g(1) < 0, and, by IVT, there

exists c in (0, 1) such that g(c)= 0, i.e., f (c)= c.

33. The domain of an even function is symmetric about the

y-axis. Since f is continuous on the right at x = 0,

therefore it must be defined on an interval [0, h] for

some h > 0. Being even, f must therefore be defined

on [−h, h]. If x = −y, then

lim

x→0−

f (x)= lim

y→0+

f (−y)= lim

y→0+

f (y)= f (0).

Thus, f is continuous on the left at x = 0. Being contin-

uous on both sides, it is therefore continuous.

34. f odd ⇔ f (−x)= − f (x)

f continuous on the right ⇔ lim

x→0+

Therefore, letting t = −x, we obtain

f (x)= f (0)

lim

x→0−

f (x)= lim

f (−t)= lim

t→0+

t→0+− f (t)

= − f (0)= f (−0)= f (0).

Therefore f is continuous at 0 and f (0)= 0.

35. max 1.593 at−0.831, min−0.756 at 0.629

36. max 0.133 at x = 1.437; min−0.232 at x = −1.805

37. max 10.333 at x = 3; min 4.762 at x = 1.260

38. max 1.510 at x = 0.465; min 0 at x = 0 and x = 1

39. root x = 0.682

40. root x = 0.739

41. roots x = −0.637 and x = 1.410

42. roots x = −0.7244919590 and x = 1.220744085

43. fsolve gives an approximation to the single real root to

10 significant figures; solve gives the three roots (includ-

ing a complex conjugate pair) in exact form involving the

quantity 108 + 12√69

1/3

; evalf(solve) gives approxi-

mations to the three roots using 10 significant figures for

the real and imaginary parts.

The Formal Definition of Limit

Section 1.5 (page 92)

1. We require 39.9 ≤ L ≤ 40.1. Thus

39.9 ≤ 39.6 + 0.025T ≤ 40.1

0.3 ≤ 0.025T ≤ 0.5

12 ≤ T ≤ 20.

The temperature should be kept between 12◦C and 20 ◦C.

2. Since 1.2% of 8,000 is 96, we require the edge length x

of the cube to satisfy 7904 ≤ x3 ≤ 8096. It is sufficient

that 19.920 ≤ x ≤ 20.079. The edge of the cube must be

within 0.079 cm of 20 cm.

3. 3− 0.02 ≤ 2x− 1 ≤ 3 + 0.02

3.98 ≤ 2x ≤ 4.02

1.99 ≤ x ≤ 2.01

4. 4− 0.1 ≤ x2 ≤ 4 + 0.1

1.9749 ≤ x ≤ 2.0024

5. 1− 0.1 ≤ √x ≤ 1.1

0.81 ≤ x ≤ 1.21

1

6.−2− 0.01 ≤

≤ −2 + 0.01

x

1

1

−

2.01 ≥ x ≥ −

1.99

−0.5025 ≤ x ≤ −0.4975

7. We need−0.03 ≤ (3x +1)−7 ≤ 0.03, which is equivalent

to−0.01 ≤ x− 2 ≤ 0.01 Thus δ= 0.01 will do.

8. We need−0.01 ≤ √2x + 3− 3 ≤ 0.01. Thus

2.99 ≤ √2x + 3 ≤ 3.01

8.9401 ≤ 2x + 3 ≤ 9.0601

2.97005 ≤ x ≤ 3.03005

3− 0.02995 ≤ x− 3 ≤ 0.03005.

Here δ= 0.02995 will do.

9. We need 8− 0.2 ≤ x3 ≤ 8.2, or 1.9832 ≤ x ≤ 2.0165.

Thus, we need−0.0168 ≤ x− 2 ≤ 0.0165. Here

δ= 0.0165 will do.

33

Copyright © 2014 Pearson Canada Inc.SECTION 1.5 (PAGE 92) ADAMS and ESSEX: CALCULUS 8

10. We need 1− 0.05 ≤ 1/(x + 1) ≤ 1 + 0.05,

or 1.0526 ≥ x + 1 ≥ 0.9524. This will occur if

−0.0476 ≤ x ≤ 0.0526. In this case we can take

δ= 0.0476.

11. To be proved: lim

x→1(3x + 1)= 4.

Proof: Let ǫ > 0 be given. Then |(3x + 1)− 4| < ǫ holds

if 3|x−1| < ǫ, and so if |x−1| < δ= ǫ/3. This confirms

the limit.

12. To be proved: lim

x→2(5− 2x)= 1.

Proof: Let ǫ > 0 be given. Then |(5− 2x)− 1| < ǫ holds

if |2x−4| < ǫ, and so if |x−2| < δ= ǫ/2. This confirms

the limit.

13. To be proved: lim

x2

= 0.

x→0

Let ǫ > 0 be given. Then |x2

− 0| < ǫ holds if

|x− 0| = |x| < δ= √ǫ.

x− 2

14. To be proved: lim

= 0.

x→2

1 + x2

Proof: Let ǫ > 0 be given. Then

x− 2

1 + x2− 0=

|x− 2|

1 + x2 ≤ |x− 2| < ǫ

provided |x− 2| < δ= ǫ.

1− 4x2

15. To be proved: lim

= 2.

x→1/2

1− 2x

Proof: Let ǫ > 0 be given. Then if x ̸= 1/2 we have

1− 4x2

1− 2x− 2 = |(1+ 2x)− 2| = |2x− 1| = 2 x−

1

2

provided |x−

1

2 | < δ= ǫ/2.

x2 + 2x

16. To be proved: lim

x→−2

x + 2= −2.

Proof: Let ǫ > 0 be given. For x ̸= −2 we have

x2 + 2x

x + 2− (−2)= |x + 2| < ǫ

provided |x + 2| < δ= ǫ. This completes the proof.

1

1

17. To be proved: lim

x→1

x + 1=

.

2

Proof: Let ǫ > 0 be given. We have

1

1

x + 1−

2=

1− x

2(x + 1)

=

|x− 1|

2|x + 1|.

If |x− 1| < 1, then 0 < x < 2 and 1 < x + 1 < 3, so that

|x + 1| > 1. Let δ= min(1, 2ǫ). If |x− 1| < δ, then

1

1

x + 1−

2=

|x− 1|

2ǫ

<

2|x + 1|

2= ǫ.

34

< ǫ

This establishes the required limit.

x + 1

1

18. To be proved: lim

x→−1

x2

− 1= −

.

2

Proof: Let ǫ > 0 be given. If x ̸= −1, we have

x + 1

x2

− 1− −

1

2=

1

x− 1− −

1

2=

|x + 1|

2|x− 1|.

If |x +1| < 1, then−2 < x < 0, so−3 < x−1 <−1 and

|x− 1| > 1. Ler δ= min(1, 2ǫ). If 0 < |x− (−1)| < δ

then |x− 1| > 1 and |x + 1| < 2ǫ. Thus

x + 1

x2

− 1− −

1

2=

|x + 1|

2ǫ

<

2|x− 1|

2= ǫ.

This completes the required proof.

19. To be proved: lim

√x = 1.

x→1

Proof: Let ǫ > 0 be given. We have

|√x− 1| =

x− 1

√x + 1 ≤ |x− 1| < ǫ

20. To be proved: lim

provided |x− 1| < δ= ǫ. This completes the proof.

x3

= 8.

x→2

Proof: Let ǫ > 0 be given. We have

|x3

− 8| = |x− 2||x2 + 2x + 4|. If |x− 2| < 1,

then 1 < x < 3 and x2 < 9. Therefore

|x2 + 2x + 4| ≤ 9 + 2 × 3 + 4= 19. If

|x− 2| < δ= min(1, ǫ/19), then

|x3

− 8| = |x− 2||x2 + 2x + 4| <

ǫ

19

× 19= ǫ.

This completes the proof.

21. We say that limx→a− f (x)= L if the following condition

holds: for every number ǫ > 0 there exists a number

δ > 0, depending on ǫ, such that

a− δ < x < a implies | f (x)− L| < ǫ.

22. We say that limx→−∞ f (x)= L if the following condi-

tion holds: for every number ǫ > 0 there exists a number

R > 0, depending on ǫ, such that

x <−R implies | f (x)− L| < ǫ.

23. We say that limx→a f (x) = −∞ if the following con-

dition holds: for every number B > 0 there exists a

number δ > 0, depending on B, such that

0 < |x− a| < δ implies f (x) <−B.

Copyright © 2014 Pearson Canada Inc.INSTRUCTOR’S SOLUTIONS MANUAL SECTION 1.5 (PAGE 92)

24. We say that limx→∞ f (x) = ∞ if the following condition

holds: for every number B > 0 there exists a number

R > 0, depending on B, such that

x > R implies f (x) > B.

25. We say that limx→a+ f (x) = −∞ if the following con-

dition holds: for every number B > 0 there exists a

number δ > 0, depending on R, such that

a < x < a + δ implies f (x) <−B.

26. We say that limx→a− f (x) = ∞ if the following con-

dition holds: for every number B > 0 there exists a

number δ > 0, depending on B, such that

a− δ < x < a implies f (x) > B.

27. To be proved: limx→1+

1

x− 1

= ∞. Proof: Let B > 0

1

be given. We have

> B if 0 < x− 1 < 1/B, that

x− 1

is, if 1 < x < 1 + δ, where δ= 1/B. This completes the

proof.

28. To be proved: limx→1−

1

x− 1

= −∞. Proof: Let B > 0

1

be given. We have

<−B if 0 > x− 1 >−1/B,

x− 1

that is, if 1− δ < x < 1, where δ= 1/B.. This completes

the proof.

1

29. To be proved: limx→∞

√x2 + 1= 0. Proof: Let ǫ > 0

be given. We have

1

√x2 + 1=

1

√x2 + 1

1

<

< ǫ

x

provided x > R, where R= 1/ǫ. This completes the

proof.

30. To be proved: limx→∞ √x = ∞. Proof: Let B > 0 be

given. We have √x > B if x > R where R= B2. This

completes the proof.

31. To be proved: if lim

x→a

f (x)= L and lim

x→a

f (x)= M, then

L= M.

Proof: Suppose L ̸= M. Let ǫ = |L− M|/3. Then

ǫ > 0. Since lim

f (x)= L, there exists δ1 > 0 such that

x→a

| f (x)−L| < ǫ if |x−a| < δ1. Since lim

f (x)= M, there

x→a

exists δ2 > 0 such that | f (x)− M| < ǫ if |x− a| < δ2.

Let δ= min(δ1, δ2). If |x− a| < δ, then

3ǫ = |L− M| = |( f (x)− M) + (L− f (x)|

≤ | f (x)− M| + | f (x)− L| < ǫ + ǫ = 2ǫ.

This implies that 3 < 2, a contradiction. Thus the origi-

nal assumption that L ̸= M must be incorrect. Therefore

L= M.

32. To be proved: if lim

g(x)= M, then there exists δ > 0

x→a

such that if 0 < |x− a| < δ, then |g(x)| < 1 + |M|.

Proof: Taking ǫ = 1 in the definition of limit, we obtain

a number δ > 0 such that if 0 < |x− a| < δ, then

|g(x)− M| < 1. It follows from this latter inequality that

|g(x)| = |(g(x)− M) + M| ≤ |G(x)− M| + |M| < 1+ |M|.

33. To be proved: if lim

x→a

f (x)= L and lim

x→a

g(x)= M, then

lim

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

x→a

Proof: Let ǫ > 0 be given. Since lim

f (x)= L, there

x→a

exists δ1 > 0 such that | f (x)− L| < ǫ/(2(1 + |M|))

if 0 < |x− a| < δ1. Since lim

g(x)= M, there ex-

x→a

ists δ2 > 0 such that |g(x)− M| < ǫ/(2(1 + |L|)) if

0 < |x− a| < δ2. By Exercise 32, there exists δ3 > 0

such that |g(x)| < 1 + |M| if 0 < |x− a| < δ3. Let

δ= min(δ1, δ2, δ3). If |x− a| < δ, then

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

= |( f (x)− L)g(x) + L(g(x)− M)|

≤ |( f (x)− L)g(x)| + |L(g(x)− M)|

= | f (x)− L||g(x)| + |L||g(x)− M|

ǫ

<

2(1 + |M|) (1 + |M|) + |L|

ǫ

2(1 + |L|)

ǫ

≤

ǫ

2 +

2= ǫ.

Thus lim

x→a

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

34. To be proved: if lim

g(x)= M where M ̸= 0, then

x→a

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

|g(x)| > |M|/2.

Proof: By the definition of limit, there exists δ > 0 such

that if 0 < |x− a| < δ, then |g(x)− M| < |M|/2

(since |M|/2 is a positive number). This latter inequality

implies that

|M| = |g(x)+(M−g(x))| ≤ |g(x)|+|g(x)−M| < |g(x)|+ |M|

2.

It follows that |g(x)| > |M| − (|M|/2)= |M|/2, as

required.

35

Copyright © 2014 Pearson Canada Inc.SECTION 1.5 (PAGE 92) ADAMS and ESSEX: CALCULUS 8

35. To be proved: if lim

g(x)= M where M ̸= 0, then

x→a

1

1

lim

=

x→a

g(x)

M.

Proof: Let ǫ > 0 be given. Since lim

g(x)= M ̸= 0,

x→a

there exists δ1 > 0 such that |g(x)− M| < ǫ|M|2/2 if

0 < |x− a| < δ1. By Exercise 34, there exists δ2 > 0

such that |g(x)| > |M|/2 if 0 < |x− a| < δ3. Let

δ= min(δ1, δ2). If 0 < |x− a| < δ, then

1

1

−

g(x)

M=

|M− g(x)|

ǫ|M|2

<

|M||g(x)|

2

2

|M|2

= ǫ.

This completes the proof.

36. To be proved: if lim

f (x)= L and lim

x→a

x→a

f (x)

L

then lim

=

x→a

g(x)

M.

Proof: By Exercises 33 and 35 we have

f (x)= M ̸= 0,

lim

x→a

f (x)

g(x)

= lim

x→a

f (x) ×

1

g(x)

L

= L ×

1

M=

M.

37. To be proved: if f is continuous at L and lim

g(x)= L,

x→c

then lim

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

x→c

Proof: Let ǫ > 0 be given. Since f is continuous at L,

there exists a number γ > 0 such that if |y− L| < γ , then

| f (y)− f (L)| < ǫ. Since limx→c g(x)= L, there exists

δ > 0 such that if 0 < |x− c| < δ, then |g(x)− L| < γ.

Taking y= g(x), it follows that if 0 < |x− c| < δ, then

| f (g(x))− f (L)| < ǫ, so that limx→c f (g(x))= f (L).

38. To be proved: if f (x) ≤ g(x) ≤ h(x) in an open interval

containing x = a (say, for a− δ1 < x < a + δ1, where

δ1 > 0), and if limx→a f (x)= limx→a h(x)= L, then

also limx→a g(x)= L.

Proof: Let ǫ > 0 be given. Since limx→a f (x)= L,

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

then | f (x)− L| < ǫ/3. Since limx→a h(x)= L,

there exists δ3 > 0 such that if 0 < |x− a| < δ3,

then |h(x)− L| < ǫ/3. Let δ= min(δ1, δ2, δ3). If

0 < |x− a| < δ, then

|g(x)− L| = |g(x)− f (x) + f (x)− L|

≤ |g(x)− f (x)| + | f (x)− L|

≤ |h(x)− f (x)| + | f (x)− L|

= |h(x)− L + L− f (x)| + | f (x)− L|

≤ |h(x)− L| + | f (x)− L| + | f (x)− L|

ǫ

ǫ

ǫ

<

3 +

3 +

3= ǫ.

Thus limx→a g(x)= L.

2

Review Exercises 1 (page 93)

1. The average rate of change of x3 over [1, 3] is

33

− 13

3− 1=

26

2= 13.

2. The average rate of change of 1/x over [−2,−1] is

(1/(−1))− (1/(−2))

−1− (−2)

=

−1/2

1= −

1

2.

3. The rate of change of x3 at x = 2 is

lim

h→0

(2 + h)3

− 23

h= lim

h→0

8 + 12h + 6h2 + h3

h

= lim

h→0(12 + 6h + h2)= 12.

− 8

4. The rate of change of 1/x at x = −3/2 is

lim

h→0

1

−(3/2) + h−

1

−3/2

h= lim

h→0

2

2h− 3 +

3

= lim

h→0

= lim

h→0

5. lim

x→1(x2

− 4x + 7)= 1− 4 + 7= 4

6. lim

x→2

x2

1− x2

22

=

= −

1− 22

4

3

7. lim

x→1

x2

1− x2 does not exist. The denominator approaches

0 (from both sides) while the numerator does not.

8. lim

x→2

x2

x2

− 4

− 5x + 6= lim

x→2

(x− 2)(x + 2)

(x− 2)(x− 3)

= lim

x→2

x + 2

x− 3= −4

x2

− 4

(x− 2)(x + 2)

9. lim

x→2

x2

− 4x + 4= lim

= lim

x→2

(x− 2)2

x→2

x + 2

x− 2

does not exist. The denominator approaches 0 (from both

sides) while the numerator does not.

h

2(3 + 2h− 3)

3(2h− 3)h

4

= −

4

3(2h− 3)

9.

36

Copyright © 2014 Pearson Canada Inc.INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 1 (PAGE 94)

1

27. lim

x→0

x sin

x

√x + 3h + √x

3=

2√x

3

10. lim

x→2−

x2

x2

− 4

− 4x + 4= lim

x→2−

x + 2

x− 2

= −∞

11. lim

x→−2+

x2

− 4

x2 + 4x + 4= x→−2+

lim

x− 2

= −∞

x + 2

2− √x

12. lim

x→4

x− 4= lim

x→4

4− x

(2 + √x)(x− 4)

1

= −

4

x2

− 9

(x− 3)(x + 3)(√x + √3)

13. lim

x→3

√x− √3= lim

x→3

x− 3

= lim

x→3(x + 3)(√x + √3)= 12√3

14. lim

h→0

h

h(√x + 3h + √x)

= lim

√x + 3h− √x

h→0

(x + 3h)− x

= lim

h→0

15. lim

x→0+

x− x2 = 0

16. lim

x→0

x− x2 does not exist because √x− x2 is not de-

fined for x < 0.

17. lim

x→1

x− x2 does not exist because √x− x2 is not de-

fined for x > 1.

18. lim

x→1−

x− x2 = 0

19. lim

x→∞

1− x2

3x2

− x− 1= lim

x→∞

(1/x2)− 1

3− (1/x)− (1/x2)

= −

3

20. lim

x→−∞

2x + 100

x2 + 3= lim

x→−∞

(2/x) + (100/x2)

1 + (3/x2)

= 0

21. lim

x→−∞

x3

− 1

x2 + 4= lim

x→−∞

x4

22. lim

x→∞

x2

− 4= lim

x→∞

x− (1/x2)

1 + (4/x2)

x2

1− (4/x2)

= −∞

= ∞

23. lim

x→0+

24. lim

x→1/2

1

√x− x2

1

√x− x2

= ∞

1

=

√1/4= 2

25. lim

sin x does not exist; sin x takes the values−1 and 1

x→∞

in any interval (R, ∞), and limits, if they exist, must be

unique.

cos x

26. lim

x→∞

= 0 by the squeeze theorem, since

x

−

1

x

≤

cos x

x

≤

1

x

for all x > 0

and limx→∞(−1/x)= limx→∞(1/x)= 0.

1

= 0 by the squeeze theorem, since

1

−|x| ≤ x sin

≤ |x| for all x ̸= 0

x

and limx→0(−|x|)= limx→0 |x| = 0.

1

28. lim

sin

x→0

x2 does not exist; sin(1/x2) takes the values−1

and 1 in any interval (−δ, δ), where δ > 0, and limits, if

they exist, must be unique.

29. lim

x→−∞

[x + x2

− 4x + 1]

x2

− (x2

− 4x + 1)

= lim

x→−∞

x− √x2

− 4x + 1

4x− 1

= lim

x→−∞

x − |x| 1− (4/x) + (1/x2)

x[4− (1/x)]

= lim

x→−∞

x + x 1− (4/x) + (1/x2)

4− (1/x)

= lim

= 2.

x→−∞

1 + 1− (4/x) + (1/x2)

Note how we have used |x| = −x (in the second last

line), because x → −∞.

30. lim

[x + x2

− 4x + 1] = ∞ + ∞ = ∞

x→∞

31. f (x)= x3

− 4x2 + 1 is continuous on the whole real line

and so is discontinuous nowhere.

x

32. f (x)=

x + 1 is continuous everywhere on its domain,

which consists of all real numbers except x = −1. It is

discontinuous nowhere.

33. f (x)=

x2 if x > 2

x if x ≤ 2 is defined everywhere and dis-

continuous at x = 2, where it is, however, left continuous

since limx→2− f (x)= 2= f (2).

34. f (x)=

x2 if x > 1

is defined and continuous ev-

x if x ≤ 1

erywhere, and so discontinuous nowhere. Observe that

limx→1− f (x)= 1= limx→1+ f (x).

35. f (x)= H(x− 1)=

1 if x ≥ 1

0 if x < 1 is defined everywhere

and discontinuous at x = 1 where it is, however, right

continuous.

36. f (x)= H(9− x2)=

1 if−3 ≤ x ≤ 3

0 if x <−3 or x > 3 is defined

everywhere and discontinuous at x = ±3. It is right

continuous at−3 and left continuous at 3.

37. f (x) = |x|+|x +1| is defined and continuous everywhere.

It is discontinuous nowhere.

38. f (x)=

|x|/|x + 1| if x ̸= −1

1 if x = −1 is defined everywhere

and discontinuous at x = −1 where it is neither left nor

right continuous since limx→−1 f (x) = ∞, while

f (−1)= 1.

37

Copyright © 2014 Pearson Canada Inc.CHALLENGING PROBLEMS 1 (PAGE 94) Challenging Problems 1 (page 94)

1. Let 0 < a < b. The average rate of change of x3 over

[a, b] is

b3

− a3

b− a

= b2 + ab + a2

.

The instantaneous rate of change of x3 at x = c is

lim

h→0

(c + h)3

− c3

h= lim

h→0

3c2h + 3ch2 + h3

h= 3c2

.

If c = (a2 + ab + b2)/3, then 3c2

= a2 + ab + b2, so

the average rate of change over [a, b] is the instantaneous

rate of change at (a2 + ab + b2)/3.

Claim: (a2 + ab + b2)/3 > (a + b)/2.

Proof: Since a2

− 2ab + b2

= (a− b)2 > 0, we have

4a2 + 4ab + 4b2 > 3a2 + 6ab + 3b2

a2 + ab + b2

a2 + 2ab + b2

>

3

4=

a + b

2

2

a2 + ab + b2

a + b

>

3

2.

2. For x near 0 we have |x− 1| = 1− x and |x + 1| = x + 1.

Thus

lim

x→0

x

|x− 1| − |x + 1|

= lim

x→0

x

(1− x)− (x + 1)

= −

1

2.

3. For x near 3 we have |5− 2x| = 2x− 5, |x− 2| = x− 2,

|x− 5| = 5− x, and |3x− 7| = 3x− 7. Thus

lim

x→3

|5− 2x| − |x− 2|

|x− 5| − |3x− 7|

= lim

x→3

= lim

x→3

2x− 5− (x− 2)

5− x− (3x− 7)

x− 3

1

= −

4(3− x)

4.

ADAMS and ESSEX: CALCULUS 8

a3

− b3

5. Use a− b=

a2 + ab + b2 to handle the denominator.

We have

lim

x→1

= lim

x→1

= lim

x→1

√3 + x− 2

3 √7 + x− 2

3 + x− 4

(7 + x)2/3 + 2(7 + x)1/3 + 4

×

√3 + x + 2

(7 + x)− 8

(7 + x)2/3 + 2(7 + x)1/3 + 4

√3 + x + 2=

4 + 4 + 4

2 + 2= 3.

6. r+(a)=

−1 + √1 + a

, r−(a)=

a

−1− √1 + a

a.

a) lima→0 r−(a) does not exist. Observe that the right

limit is −∞ and the left limit is ∞.

b) From the following table it appears that

lima→0 r+(a)= 1/2, the solution of the linear equa-

tion 2x− 1= 0 which results from setting a = 0 in

the quadratic equation ax2 + 2x− 1= 0.

a r+(a)

1 0.41421

0.1 0.48810

−0.1 0.51317

0.01 0.49876

−0.01 0.50126

0.001 0.49988

−0.001 0.50013

4. Let y= x1/6. Then we have

lim

x→64

x1/3

x1/2

− 4

− 8= lim

y→2

= lim

y→2

y + 2

y2 + 2y + 4=

4

12=

1

y2

− 4

y3

− 8

(y− 2)(y + 2)

(y− 2)(y2 + 2y + 4)

= lim

y→2

38

3.

√1 + a− 1

c) lim

a→0

r+(a)= lim

a→0

= lim

a→0

= lim

a→0

a

(1 + a)− 1

a(√1 + a + 1)

1

√1 + a + 1=

1

2.

7. TRUE or FALSE

a) If limx→a f (x) exists and limx→a g(x) does not

exist, then limx→a f (x) + g(x) does not exist.

TRUE, because if limx→a f (x) + g(x) were to

exist then

lim

x→a

g(x)= lim

x→a

f (x) + g(x)− f (x)

= lim

x→a

f (x) + g(x)− lim

x→a

f (x)

would also exist.

b) If neither limx→a f (x) nor limx→a g(x) exists, then

limx→a f (x) + g(x) does not exist.

FALSE. Neither limx→0 1/x nor limx→0(−1/x) ex-

ist, but limx→0 (1/x) + (−1/x)= limx→0 0= 0

exists.

Copyright © 2014 Pearson Canada Inc.INSTRUCTOR’S SOLUTIONS MANUAL CHALLENGING PROBLEMS 1 (PAGE 94)

c) If f is continuous at a, then so is | f |.

TRUE. For any two real numbers u and v we have

|u| − |v| ≤ |u− v|.

This follows from

|u| = |u− v + v| ≤ |u− v| + |v|, and

|v| = |v− u + u| ≤ |v− u| + |u| = |u− v| + |u|.

Now we have

| f (x)| − | f (a)| ≤ | f (x)− f (a)|

so the left side approaches zero whenever the right

side does. This happens when x → a by the conti-

nuity of f at a.

d) If | f | is continuous at a, then so is f.

FALSE. The function f (x)=

−1 if x < 0

1 if x ≥ 0 is

discontinuous at x = 0, but | f (x)| = 1 everywhere,

and so is continuous at x = 0.

e) If f (x) < g(x) in an interval around a and if

limx→a f (x)= L and limx→a g(x)= M both

exist, then L < M.

FALSE. Let g(x)=

x2 if x ̸= 0

and let

1 if x = 0

f (x)= −g(x). Then f (x) < g(x) for all x, but

limx→0 f (x)= 0= limx→0 g(x). (Note: under the

given conditions, it is TRUE that L ≤ M, but not

necessarily true that L < M.)

8. a) To be proved: if f is a continuous function defined

on a closed interval [a, b], then the range of f is a

closed interval.

Proof: By the Max-Min Theorem there exist num-

bers u and v in [a, b] such that f (u) ≤ f (x) ≤ f (v)

for all x in [a, b]. By the Intermediate-Value The-

orem, f (x) takes on all values between f (u)

and f (v) at values of x between u and v, and

hence at points of [a, b]. Thus the range of f is

[ f (u), f (v)], a closed interval.

b) If the domain of the continuous function f is an

open interval, the range of f can be any interval

(open, closed, half open, finite, or infinite).

1

2 ].

x2

− 1

9. f (x)=

|x2

− 1|

=

−1 if−1 < x < 1

1 if x <−1 or x > 1 .

f is continuous wherever it is defined, that is at all

points except x = ±1. f has left and right limits−1

and 1, respectively, at x = 1, and has left and right limits

1 and−1, respectively, at x = −1. It is not, however,

discontinuous at any point, since−1 and 1 are not in its

domain.

1

1

1

10. f (x)=

=

=

x− x2

1

4−

1

4− x + x2

1

4− x−

Observe that f (x) ≥ f (1/2)= 4 for all x in (0, 1).

1

2.

2

11. Suppose f is continuous on [0, 1] and f (0)= f (1).

a) To be proved: f (a)= f (a + 1

2 ) for some a in [0,

Proof: If f (1/2)= f (0) we can take a = 0 and be

done. If not, let

g(x)= f (x + 1

2 )− f (x).

Then g(0) ̸= 0 and

g(1/2)= f (1)− f (1/2)= f (0)− f (1/2)= −g(0).

Since g is continuous and has opposite signs at

x = 0 and x = 1/2, the Intermediate-Value The-

orem assures us that there exists a between 0 and

1/2 such that g(a)= 0, that is, f (a)= f (a + 1

2 ).

b) To be proved: if n > 2 is an integer, then

f (a)= f (a + 1

n ) for some a in [0, 1−

1

n ].

Proof: Let g(x)= f (x + 1

n )− f (x). Consider

the numbers x = 0, x = 1/n, x = 2/n, . . . ,

x = (n− 1)/n. If g(x)= 0 for any of these num-

bers, then we can let a be that number. Otherwise,

g(x) ̸= 0 at any of these numbers. Suppose that the

values of g at all these numbers has the same sign

(say positive). Then we have

f (1) > f (n−1

n ) >· · · > f ( 2

n ) > 1

n

> f (0),

which is a contradiction, since f (0)= f (1). There-

fore there exists j in the set {0, 1, 2, . . . , n− 1} such

that g( j/n) and g(( j + 1)/n) have opposite sign. By

the Intermediate-Value Theorem, g(a)= 0 for some

a between j/n and ( j + 1)/n, which is what we had

to prove.