Solution manual calculus 8th edition varberg, purcell, rigdon ch01

1

CHAPTER

Limits
9.

1.1 Concepts Review

x3 – 4 x 2 + x + 6
x → –1
x +1
lim

( x + 1)( x 2 – 5 x + 6)
x → –1
x +1

1. L; c

= lim

2. 6

= lim ( x 2 – 5 x + 6)
x → –1
2

3. L; right

= (–1) – 5(–1) + 6

4. lim f ( x) = M

= 12

x →c

Problem Set 1.1

x2
= lim( x 2 + 2 x –1) = –1
x →0

1. lim( x – 5) = –2

x →0

x →3

2. lim (1 – 2t ) = 3

11.

t → –1

3.
4.

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

= –t – t = –2t

lim ( x 2 + 2t − 1) = (−2) 2 + 2t − 1 = 3 + 2t

x →−2

(

2

) ( ( −1)

6. lim t 2 − x 2 =
t →−1

12.

) ( ( −1) − 1) = 0

5. lim t 2 − 1 =
t →−1

2

)

x2 – 4
( x – 2)( x + 2)
= lim
x→2 x – 2

x→2
x–2
= lim( x + 2)

x2 – 9
x →3 x – 3
( x – 3)( x + 3)
= lim
x →3
x–3
= lim( x + 3)
lim

x →3

− x2 = 1 − x2

7. lim

=3+3=6
13.

x→2

lim

(t + 4)(t − 2) 4
(3t − 6) 2

t →2

= lim

=2+2=4
8.

x2 – t 2
( x + t )( x – t )
= lim
x→–t x + t
x→ – t
x+t
= lim ( x – t )
lim

x→ –t

x →−2

(

x 4 + 2 x3 – x 2

10. lim

(t − 2) 2 t + 4
9(t − 2) 2

t →2

t 2 + 4t – 21
t → –7
t+7
(t + 7)(t – 3)
= lim
t → –7
t+7
= lim (t – 3)

t+4
9

= lim

lim

t →2

=

2+4
6
=
9
9

t → –7

= –7 – 3 = –10

14.

(t − 7)3
t −7

lim

t →7+

= lim
t →7

+

= lim

t →7+

(t − 7) t − 7
t −7
t −7

= 7−7 = 0

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63

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or by any means, without permission in writing from the publisher.


15. lim

x 4 –18 x 2 + 81

x →3

( x – 3)

2

= lim

x →3

( x – 3) 2 ( x + 3) 2

= lim

x →3

( x – 3)

( x 2 – 9) 2
( x – 3)

= lim( x + 3)2 = (3 + 3) 2

2

lim

t →0

2

(3u + 4)(2u – 2)3
(u –1) 2

u →1

= lim

( x − sin x ) 2 / x 2

x

21.

x →3

1.

0.0251314

0.1

2.775 × 10−6

8(3u + 4)(u –1)3

0.01

2.77775 × 10−10

(u –1) 2

0.001

2.77778 × 10−14

–1.
–0.1

0.0251314
2.775 × 10−6

–0.01

2.77775 × 10−10

–0.001

2.77778 × 10−14

= 36

16. lim

1 − cos t
=0
2t

u →1

= lim 8(3u + 4)(u – 1) = 8[3(1) + 4](1 – 1) = 0
u →1

17.

(2 + h) 2 − 4
4 + 4h + h 2 − 4
= lim
h→0
h→0
h
h
lim

h 2 + 4h
= lim(h + 4) = 4
h →0
h →0
h

= lim

lim

( x – sin x) 2
x2

x →0

18.

( x + h) 2 − x 2
x 2 + 2 xh + h 2 − x 2
= lim
h→0
h →0
h
h
lim

h 2 + 2 xh
= lim(h + 2 x) = 2 x
h →0
h →0
h

= lim

sin x
2x

x

19.

2

(1 − cos x ) / x

x

22.

0.211322

0.1

0.00249584

0.01
0.001

0.0000249996
2.5 × 10−7

–1.

0.211322

–0.1

0.00249584

0.0000249996
2.5 × 10−7

0.420735

0.1

0.499167

0.01

0.499992

–0.01

0.001

0.49999992

–0.001

–1.

0.420735

–0.1

0.499167

–0.01

0.499992

–0.001

0.49999992

0.01
0.001

x2

x →0

2

(t − 1) /(sin(t − 1))

t

23.

=0

1.1

2.1035

1.01

2.01003

1.001

2.001

0.229849

0

1.1884

0.0249792

0.9

1.90317

0.00249998

0.99

1.99003

0.999

1.999

1− cos t
2t

t

0.1

(1 – cos x) 2

3.56519

sin x
= 0.5
x →0 2 x

1.

lim

2.

lim

0.00024999998

2

1.

1.

20.

=0

t −1
=2
− 1)
2

lim

64

–1.

–0.229849

–0.1

–0.0249792

–0.01

–0.00249998

–0.001

–0.00024999998

Section 1.1

t →1 sin(t

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or by any means, without permission in writing from the publisher.


x −sin( x − 3) − 3
x −3

x

24.

4.

1. + π4
0.1 +

0.158529

3.1

2.

lim

x→ π
4

(1 + sin( x − 3π / 2)) /( x − π )

x
1. + π

0.4597

0.1 + π

0.0500

0.01 + π

0.0050

0.001 + π

0.0005

–0.0050

–0.001 + π

–0.0005

1 + sin ( x − 32π )
x−π

x →π

=0

−0.1 +

0.0000210862
2.12072 × 10−7

π
2

0.536908
π
2

0.00226446
π
2

π
2

0.0000213564
2.12342 × 10−7

2 − 2sin u
lim
=0
π
u→

3u
2

29. a.

–0.896664

0.01

–0.989967

0.001

–0.999

d.

–1.

–1.64209

e.

–0.1

–1.09666

f.

–0.01

–1.00997

–0.001

–1.001

lim f ( x) = 2

x → –3

b. f(–3) = 1
c.

g.

= –1
h.

i.

Instructor’s Resource Manual

0.00199339

−0.001 +

0.1

1

t

0.11921

−0.01 +

0.357907

1 – cot t

= 0.25
(2 − 2sin u ) / 3u

0.001 +

1.

t →0

(tan x − 1)2

0.01 + π2

(1 − cot t ) /(1 / t )

lim

(x − )

0.2505

0.1 + π2

t

26.

0.255008

1. + π2

–0.0500

–0.01 + π

lim

0.300668

u

28.

–0.4597

–0.1 + π

0.674117

π 2

4

−1. + π2

–1. + π

0.2495

4

−0.001 + π4

x – sin( x – 3) – 3
=0
lim
x →3
x–3
25.

0.245009
π

−0.01 + π4

0.0000166666
1.66667 × 10−7

2.999

4

−0.1 + π4

0.00166583

2.99

0.201002
π

−1. + π4

0.158529

2.9

4

0.001 +

0.0000166666
1.66667 × 10−7

3.001

0.0320244
π

0.01 +

0.00166583

3.01

( x − π / 4) 2 /(tan x − 1) 2

x

27.

f(–1) does not exist.
lim f ( x) =

x → –1

5
2

f(1) = 2
lim f(x) does not exist.
x→1

lim f ( x) = 2

x →1–

lim f ( x) = 1

x →1+

lim f ( x ) =
+

x →−1

5
2

Section 1.1

65

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or by any means, without permission in writing from the publisher.


lim f ( x) does not exist.

b.

lim f ( x) does not exist.

b.

f(–3) = 1

c.

f(1) = 2

c.

f(–1) = 1

d.

30. a.

d.

x → –3

lim f ( x) = 2

x → –1

e.

f(1) = 1

f.

lim f ( x) does not exist.

g.
h.
i.
31. a.
b.
c.

d.
e.
f.
32. a.
b.
c.

x →1

lim f ( x) = 2

x →1+

34.

x →1

lim f ( x) = 1

x →1–

lim f ( x) does not exist.

x →1+

lim f ( x ) = 2

x →−1+

a.

f(–3) = 2
f(3) is undefined.

x →1

b. g(1) does not exist.

lim f ( x) = 2

c.

x → –3−

lim f ( x) = 4

x → –3+

d.

lim f ( x) does not exist.

x → –3

lim g ( x) = 0

35.

lim g ( x ) = 1

x→2

lim g ( x ) = 1

x → 2+

f ( x) = x – ⎣⎡[ x ]⎦⎤

lim f ( x) does not exist.

x →3+

lim f ( x) = −2

x → –1−

lim f ( x) = −2

x → –1+

lim f ( x) = −2

x → –1

d. f (–1) = –2
e.

lim f ( x) = 0

f.

f (1) = 0

x →1

a.
b.

33.
c.

d.

a.

66

f(0) = 0
lim f ( x) does not exist.

x →0

lim f ( x ) = 1

x →0 –

lim f ( x) =

x→ 1
2

1
2

lim f ( x) = 0

x →0

Section 1.1

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or by any means, without permission in writing from the publisher.


f ( x) =

36.

41. lim f ( x) exists for a = –1, 0, 1.

x
x

x→a

42. The changed values will not change lim f ( x) at
x→a

any a. As x approaches a, the limit is still a 2 .
43. a.

x −1

lim

x →1

lim

x −1
x −1

−

x →1

b.

f (0) does not exist.

a.

lim f ( x) does not exist.

b.

x →0

lim

−

x →1

lim

c.

x −1
x −1
x −1

does not exist.
= −1 and lim

+

x →1

x −1

=1

= −1

x2 − x − 1 − 1
x −1

x →1−

x −1

= −3

lim f ( x ) = –1

c.

x →0 –

d.

⎡ 1
1 ⎤
lim ⎢
−
⎥ does not exist.
− x −1
x − 1 ⎥⎦
x →1 ⎢
⎣

d.

lim f ( x) = 1

x→ 1

2

44. a.

x2 − 1
37. lim
does not exist.
x →1 x − 1
lim

x →1−

x →0

= lim

c1f
lim dd gg does not exist.
+ x
x →0 e h

x+2− 2
x

c.

lim x(−1)ed

( x + 2 − 2)( x + 2 + 2)

d.

x →0

x+2−2
x( x + 2 + 2)

= lim

x →0

39. a.
b.

c1/ x f
hg

x →0

= lim

x →0

1

x

2
=
=

=
4
0+2 + 2 2 2
x+2+ 2

=0

c1/ x f
hg

x →0

+

45. a) 1

x( x + 2 + 2)

1

+

lim a x b (−1)ed

x( x + 2 + 2)

x →0

= lim

x − a xb = 0

b.

x2 − 1
x2 − 1
=2
= −2 and lim
x −1
x →1+ x − 1

38. lim

lim
x →1+

b) 0

−1

c)

=0

d)

−1

1

46. a) Does not exist
c)

lim f ( x) does not exist.

1

b) 0
d) 0.556

x →1

lim f ( x) = 0

47. lim x does not exist since
x →0

x →0

40.

x is not defined

for x < 0.
48.

lim x x = 1

x → 0+

49. lim

x →0

x =0
x

50. lim x = 1
x →0

sin 2 x 1
=
x →0 4 x
2

51. lim

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67

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or by any means, without permission in writing from the publisher.


52. lim

x →0

7. If x is within 0.001 of 2, then 2x is within 0.002
of 4.

sin 5 x 5
=
3x
3

⎛1⎞
53. lim cos ⎜ ⎟ does not exist.
x →0
⎝ x⎠
⎛1⎞
54. lim x cos ⎜ ⎟ = 0
x →0
⎝ x⎠

x3 − 1

55. lim

56. lim

x →0

57.

=6

2x + 2 − 2

x →1

x sin 2 x
sin( x 2 )

lim

x →2–

58. lim

+

x →1

8. If x is within 0.0005 of 2, then x2 is within 0.002
of 4.

=2

x2 – x – 2
= –3
x–2

2
1/( x −1)

1+ 2

=0

59. lim x ; The computer gives a value of 0, but
x →0

lim

x →0−

9. If x is within 0.0019 of 2, then
0.002 of 4.

8 x is within

x does not exist.

1.2 Concepts Review
1. L – ε ; L + ε
2. 0 < x – a < δ ; f ( x) – L < ε
10. If x is within 0.001 of 2, then
3.

ε

8
is within 0.002
x

of 4.

3

4. ma + b

Problem Set 1.2
1. 0 < t – a < δ ⇒ f (t ) – M < ε
2. 0 < u – b < δ ⇒ g (u ) – L < ε

2x – 1+ 1 < ε ⇔ 2x < ε

3. 0 < z – d < δ ⇒ h( z ) – P < ε

⇔ 2 x <ε

4. 0 < y – e < δ ⇒ φ ( y ) – B < ε

⇔ x <

5. 0 < c – x < δ ⇒ f ( x) – L < ε
6. 0 < t – a < δ ⇒ g (t ) – D < ε

68

11. 0 < x – 0 < δ ⇒ (2 x – 1) – (–1) < ε

Section 1.2

ε
2

ε
δ = ;0 < x –0 <δ

2
(2 x – 1) – (–1) = 2 x = 2 x < 2δ = ε

Instructor’s Resource Manual

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or by any means, without permission in writing from the publisher.


12. 0 < x + 21 < δ ⇒ (3x – 1) – (–64) < ε
3 x – 1 + 64 < ε ⇔ 3 x + 63 < ε

⇔ 3( x + 21) < ε

2 x 2 – 11x + 5
(2 x – 1)( x – 5)
–9 <ε ⇔
–9 <ε
x–5
x–5

⇔ 3 x + 21 < ε
⇔ x + 21 <

ε

⇔ 2x – 1 – 9 < ε

3

⇔ 2( x – 5) < ε

ε
δ = ; 0 < x + 21 < δ

3
(3 x – 1) – (–64) = 3 x + 63 = 3 x + 21 < 3δ = ε

x 2 – 25
13. 0 < x – 5 < δ ⇒
– 10 < ε
x–5
x 2 – 25
( x – 5)( x + 5)
– 10 < ε ⇔
– 10 < ε
x–5
x–5
⇔ x + 5 – 10 < ε

⇔ x–5 <

ε
2

ε

δ = ;0 < x –5 <δ
2

2 x – 11x + 5
(2 x – 1)( x – 5)
–9 =
–9
x–5
x–5
2

= 2 x – 1 – 9 = 2( x – 5) = 2 x – 5 < 2δ = ε

16. 0 < x – 1 < δ ⇒

⇔ x–5 <ε

2x – 2 < ε

2x – 2 < ε

δ = ε; 0 < x – 5 < δ

( 2 x – 2 )( 2 x + 2 )

⇔

x – 25
( x – 5)( x + 5)
– 10 =

– 10 = x + 5 – 10
x–5
x–5

2x + 2

2

2x – 2

⇔

2x + 2

= x–5 <δ =ε
2

⇔2

2x – x
14. 0 < x – 0 < δ ⇒
− (−1) < ε
x
2 x2 – x
x(2 x – 1)
+1 < ε ⇔
+1 < ε
x
x
⇔ 2x < ε

⇔ 2 x <ε

ε
2

ε
δ = ;0 < x –0 <δ
2

2 x2 – x
x(2 x – 1)
− (−1) =
+ 1 = 2x – 1+ 1
x
x

<ε

x –1
2x + 2

<ε

2ε
; 0 < x –1 < δ
2
( 2 x – 2)( 2 x + 2)
2x − 2 =
2x + 2

=

2x – 2
2x + 2
2 x –1
2 x – 1 2δ
≤
<
=ε
2x + 2
2
2

17. 0 < x – 4 < δ ⇒
2x – 1
x–3

= 2 x = 2 x < 2δ = ε
⇔
⇔
⇔

Instructor’s Resource Manual

<ε

δ=

⇔ 2x – 1 +1 < ε

⇔ x <

2 x 2 – 11x + 5
–9 <ε
x–5

15. 0 < x – 5 < δ ⇒

2x – 1
x–3

– 7 <ε ⇔

– 7 <ε

2 x – 1 – 7( x – 3)
x–3

<ε

( 2 x – 1 – 7( x – 3))( 2 x – 1 + 7( x – 3))
x – 3( 2 x – 1 + 7( x – 3))
2 x – 1 – (7 x – 21)
x – 3( 2 x – 1 + 7( x – 3))
–5( x – 4)
x – 3( 2 x – 1 + 7( x – 3))

<ε

<ε

<ε

Section 1.2

69

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or by any means, without permission in writing from the publisher.


⇔ x−4 ⋅

To bound

5
x − 3( 2 x − 1 + 7( x − 3))

x – 3( 2 x – 1 + 7( x – 3))

1
2

1
2

, agree that

7
9
< x < , so

2
2

5
x – 3( 2 x – 1 + 7( x – 3))

hence x − 4 ⋅

19. 0 < x – 1 < δ ⇒

< 1.65 and

x − 3( 2 x − 1 + 7( x − 3))

<ε

2x −1

5

− 7 = x−4 ⋅
x −3
x − 3( 2 x − 1 + 7( x − 3))
< x – 4 (1.65) < 1. 65δ ≤ ε
1
1
ε
since δ = only when ≤
so 1.65δ ≤ ε .
2

2 1. 65

14 x 2 – 20 x + 6
–8 < ε
x –1

14 x 2 – 20 x + 6
2(7 x – 3)( x – 1)
–8 <ε ⇔
–8 <ε
x –1
x –1
⇔ 2(7 x – 3) – 8 < ε
⇔ 14( x – 1) < ε
⇔ 14 x – 1 < ε

δ=

ε
14

( x – 1)2

–4 <ε

–4 <ε

⇔ 10 x – 6 – 4 < ε
⇔ 10 x – 1 < ε
⇔ x –1 <

1.65
For whatever ε is chosen, let δ be the smaller of
1
ε
and
.
1.65
2
⎧1
ε ⎫
δ = min ⎨ ,
⎬, 0 < x – 4 < δ
⎩ 2 1. 65 ⎭

⇔ x –1 <

(10 x – 6)( x – 1)2

–4 <ε

⇔ 10( x – 1) < ε

5

18. 0 < x – 1 < δ ⇒

( x – 1) 2

( x –1)2

⇔

ε

⇔ x–4 <

10 x3 – 26 x 2 + 22 x – 6

10 x3 – 26 x 2 + 22 x – 6

5

δ ≤ . If δ ≤ , then
0.65 <

<ε

ε
14

δ=

ε
10

ε
10

; 0 < x –1 < δ

10 x3 – 26 x 2 + 22 x – 6
( x – 1) 2

–4 =

(10 x – 6)( x – 1) 2
( x – 1) 2

–4

= 10 x − 6 − 4 = 10( x − 1)
= 10 x − 1 < 10δ = ε

20. 0 < x – 1 < δ ⇒ (2 x 2 + 1) – 3 < ε
2 x2 + 1 – 3 = 2 x2 – 2 = 2 x + 1 x – 1

To bound 2 x + 2 , agree that δ ≤ 1 .
x – 1 < δ implies

2x + 2 = 2x – 2 + 4
≤ 2x – 2 + 4

<2+4=6

ε
⎧ ε⎫
δ ≤ ; δ = min ⎨1, ⎬; 0 < x – 1 < δ
6

⎩ 6⎭

(2 x + 1) – 3 = 2 x 2 – 2
2

⎛ε ⎞
= 2x + 2 x −1 < 6 ⋅ ⎜ ⎟ = ε
⎝6⎠

; 0 < x –1 < δ

14 x 2 – 20 x + 6
2(7 x – 3)( x – 1)
–8 =
–8
x –1
x –1
= 2(7 x – 3) – 8
= 14( x – 1) = 14 x – 1 < 14δ = ε

70

Section 1.2

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or by any means, without permission in writing from the publisher.


21. 0 < x + 1 < δ ⇒ ( x 2 – 2 x – 1) – 2 < ε

x2 – 2 x – 1 – 2 = x2 – 2 x – 3 = x + 1 x – 3

To bound x – 3 , agree that δ ≤ 1 .
x + 1 < δ implies

⎛1⎞
25. For all x ≠ 0 , 0 ≤ sin 2 ⎜ ⎟ ≤ 1 so
⎝x⎠
1
⎛ ⎞
x 4 sin 2 ⎜ ⎟ ≤ x 4 for all x ≠ 0 . By Problem 18,
⎝ x⎠
4
lim x = 0, so, by Problem 20,
x→0

4
2 ⎛ 1⎞
lim x sin ⎜ ⎟ = 0.
⎝ x⎠
x→0

x – 3 = x + 1 – 4 ≤ x + 1 + –4 < 1 + 4 = 5

ε
⎧ ε⎫
δ ≤ ; δ = min ⎨1, ⎬ ; 0 < x + 1 < δ
⎩ 5⎭

5

26. 0 < x < δ ⇒

( x – 2 x – 1) – 2 = x 2 – 2 x – 3
2

= x +1 x – 3 < 5⋅

ε
5

x –0 =

x = x <ε

2
For x > 0, ( x ) = x.

=ε

x < ε ⇔ ( x )2 = x < ε 2

δ = ε 2; 0 < x < δ ⇒ x < δ = ε 2 = ε

22. 0 < x < δ ⇒ x 4 – 0 = x 4 < ε
x 4 = x x3 . To bound x3 , agree that
3

δ ≤ 1. x < δ ≤ 1 implies x3 = x ≤ 1 so
δ ≤ ε.

δ = min{1, ε }; 0 < x < δ ⇒ x 4 = x x3 < ε ⋅1

27.

lim x : 0 < x < δ ⇒ x – 0 < ε

x →0 +

For x ≥ 0 , x = x .

δ = ε; 0 < x < δ ⇒ x – 0 = x = x < δ = ε
Thus, lim+ x = 0.
x→0

lim x : 0 < 0 – x < δ ⇒ x – 0 < ε

=ε

23. Choose ε > 0. Then since lim f ( x) = L, there is
x →c

some δ1 > 0 such that
0 < x – c < δ1 ⇒ f ( x ) – L < ε .

x →0 –

For x < 0, x = – x; note also that
since x ≥ 0.

x = x

δ = ε ;0 < − x < δ ⇒ x = x = − x < δ = ε

Since lim f (x) = M, there is some δ 2 > 0 such

Thus, lim– x = 0,

that 0 < x − c < δ 2 ⇒ f ( x) − M < ε .

since lim x = lim x = 0, lim x = 0.

x→c

Let δ = min{δ1 , δ2 } and choose x 0 such that
0 < x0 – c < δ .
Thus, f ( x0 ) – L < ε ⇒ −ε < f ( x0 ) − L < ε
⇒ − f ( x0 ) − ε < − L < − f ( x0 ) + ε
⇒ f ( x0 ) − ε < L < f ( x0 ) + ε .
Similarly,
f ( x0 ) − ε < M < f ( x0 ) + ε .
Thus,
−2ε < L − M < 2ε . As ε ⇒ 0, L − M → 0, so
L = M.

24. Since lim G(x) = 0, then given any ε > 0, we

x→0

x →0 +

x →0

x →0 –

28. Choose ε > 0. Since lim g( x) = 0 there is some
x→ a

δ1 > 0 such that

0 < x – a < δ1 ⇒ g(x ) − 0 <

ε.
B

Let δ = min{1, δ1} , then f ( x) < B for
x − a < δ or x − a < δ ⇒ f ( x) < B. Thus,
x − a < δ ⇒ f ( x) g ( x) − 0 = f ( x) g ( x)
= f ( x) g ( x) < B ⋅

ε
B

= ε so lim f ( x)g(x) = 0.
x→ a

x→c

can find δ > 0 such that whenever
x – c < δ , G ( x) < ε .
Take any ε > 0 and the corresponding δ that

works for G(x), then x – c < δ implies
F ( x) – 0 = F ( x) ≤ G ( x ) < ε since
lim G(x) = 0.
x→c

Thus, lim F( x) = 0.
x→c

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or by any means, without permission in writing from the publisher.


29. Choose ε > 0. Since lim f ( x) = L, there is a

1.3 Concepts Review

x→ a

δ > 0 such that for 0 < x – a < δ , f ( x) – L < ε .
That is, for
a − δ < x < a or a < x < a + δ ,
L − ε < f ( x) < L + ε .
Let f(a) = A,
M = max { L − ε , L + ε , A } , c = a – δ,

d = a + δ. Then for x in (c, d), f ( x) ≤ M , since
either x = a, in which case
f ( x) = f (a ) = A ≤ M or 0 < x – a < δ so

1. 48
2. 4
3. – 8; – 4 + 5c
4. 0

Problem Set 1.3
1. lim (2 x + 1)
= lim 2 x + lim 1
x→1

x →1

α

and δ = min{δ1 , δ 2} where
2
0 < x – a < δ1 ⇒ f ( x) – L < ε and

2.

x→ –1

L – ε < f(x) < L + ε and M – ε < g(x) < M + ε.
Combine the inequalities and use the fact
that f ( x) ≤ g ( x) to get

L – ε < f(x) ≤ g(x) < M + ε which leads to
L – ε < M + ε or L – M < 2ε.
However,
L – M = α > 2ε
which is a contradiction.
Thus L ≤ M .

x+6

x – 4x + x 2 + x + 6
an asymptote at x ≈ 3.49.

c.

2

= 3(–1) – 1 = 2

3. lim [(2 x +1)( x – 3)]

+ 1 has

1
, then 2.75 < x < 3
4
or 3 < x < 3.25 and by graphing

6

x→0

= lim (2 x +1) ⋅ lim (x – 3)
x→ 0

x→ 0

⎞ ⎛
⎞
⎛
= ⎜ lim 2 x + lim 1⎟ ⋅ ⎜ lim x – lim 3⎟
⎝ x→ 0
x→ 0 ⎠ ⎝ x→0
x→ 0 ⎠
⎞ ⎛
⎞
⎛
= ⎜ 2 lim x + lim 1⎟ ⋅ ⎜ lim x – lim 3⎟
⎠
⎝ x →0
⎝
x→ 0
x→0
x→ 0 ⎠
= [2(0) +1](0 – 3) = –3
lim [(2 x 2 + 1)(7 x 2 + 13)]

x→ 2

x→ 2

If δ ≤

y = g ( x) =

2, 1

= lim (2 x 2 + 1) ⋅ lim (7 x 2 + 13)

x 4 – 4x 3 + x 2 + x + 6
3

2

8

x→–1

⎛
⎞
= 3⎜ lim x ⎟ – lim 1
⎝ x→ –1 ⎠
x →–1

4.

x 3 – x 2 – 2x – 4

3

x→–1

x→ –1

32. For every ε > 0 and δ > 0 there is some x with
0 < x – c < δ such that f ( x ) – L > ε .

4

5

= 3 lim x 2 – lim 1

31. (b) and (c) are equivalent to the definition of
limit.

b. No, because

lim (3x 2 – 1)

x→ –1

= lim 3x 2 – lim 1

Thus, for 0 < x – a < δ ,

g(x) =

2,1

x→1

= 2(1) + 1 = 3

0 < x – a < δ 2 ⇒ g ( x) – M < ε .

33. a.

3

x→1

= 2 lim x + lim 1

30. Suppose that L > M. Then L – M = α > 0. Now

take ε <

4

x→1

L − ε < f ( x) < L + ε and f ( x) < M .

x→ 2

4, 5
3
2, 1

6

4, 3

⎛
⎞ ⎛
⎞
= ⎜ 2 lim x 2 + lim 1⎟ ⋅ ⎜ 7 lim x 2 + lim 13 ⎟ 8,1
x→ 2 ⎠ ⎝ x→ 2
x→ 2 ⎠
⎝ x→ 2
2
2
⎡ ⎛
⎤
⎡
⎤
⎞
⎞
⎛
= ⎢2⎜ lim x ⎟ + 1⎥ ⎢7⎜ lim x ⎟ + 13⎥
2
⎢⎣ ⎝ x → 2 ⎠
⎥⎦ ⎢⎣ ⎝ x → 2 ⎠
⎥⎦
= [2( 2 ) 2 + 1][7( 2 ) 2 + 13] = 135

x3 − x 2 − 2 x − 4

x 4 − 4 x3 + x 2 + x + 6
on the interval [2.75, 3.25], we see that
0<

x3 – x 2 – 2 x – 4

<3
x 4 – 4 x3 + x 2 + x + 6
so m must be at least three.

72

Section 1.3

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or by any means, without permission in writing from the publisher.


2x + 1
x→2 5 – 3x
lim (2 x + 1)
= x→2
lim (5 – 3 x)

9.

7

5. lim

4, 5

x→2

=

3, 1

lim 5 – lim 3 x

x→2

13

4, 3

⎡
⎤
= ⎢2 lim t 3 + lim 15⎥
⎣ t→ –2
t→ –2 ⎦

lim 2 x + lim 1

x→2

8

t→ –2

⎡

⎤
= ⎢ lim (2t3 + 15) ⎥
⎣t→–2
⎦

x→2

=

lim (2t 3 +15)13

13

2 lim x + 1
x→2

8

3
⎡
⎤
= ⎢ 2 ⎛⎜ lim t ⎞⎟ + lim 15⎥
⎣⎢ ⎝ t → –2 ⎠ t → –2 ⎦⎥

x→2

2, 1

= [2(–2) 3 + 15]13 = –1

2

5 – 3 lim x

13

x→2

2(2) + 1
=
= –5
5 – 3(2 )

10.

=

3

6.

4x +1

lim

7

x → –3 7 – 2 x 2

lim (4 x + 1)

lim (7 – 2 x )
lim 4 x 3 + lim 1
x → –3

lim 7 – lim 2 x 2

x → –3

=

x → –3

2

x →3

13

5, 3

= 3 lim x – lim 5

2, 1

x →3

lim

x → –3

=

5x2 + 2 x

9

2

lim (5 x + 2 x )

x → –3

= 5 lim x 2 + 2 lim x
x → –3

x → –3

⎛ 4 lim y 3 + 8 lim y ⎞
⎜ y →2
y →2 ⎟
=⎜
⎟
y + lim 4 ⎟
⎜ ylim
→
y
→
2
2

⎝
⎠

8, 1
1/ 3

= 3(3) – 5 = 2

8.

4, 3
13

lim (3 x – 5)

x →3

x →3

7

⎡ lim (4 y 3 + 8 y ) ⎤
⎢ y →2
⎥
=⎢
( y + 4) ⎥⎥
⎢ ylim
→
2
⎣

⎦

9

7. lim 3 x – 5

2

9

⎛
4 y3 + 8 y ⎞
= ⎜ lim
⎟
⎜ y →2 y + 4 ⎟
⎝
⎠

3

=

⎛ 4 y3 + 8 y ⎞
lim ⎜
⎟
y →2 ⎜ y + 4 ⎟
⎝
⎠

1/ 3

x → –3

4⎛⎜ lim x ⎞⎟ + 1
x → –3 ⎠
= ⎝
2
7 – 2⎛⎜ lim x ⎞⎟
⎝ x → –3 ⎠
4(–3)3 + 1 107
=
=
11
7 – 2(–3) 2

2

1/ 3

11.

8

7 – 2 lim x 2

8

w→ –2

= –3(–2)3 + 7(–2) 2 = 2 13

3, 1

x → –3

4 lim x 3 + 1

4, 3

= –3 ⎛⎜ lim w ⎞⎟ + 7 ⎛⎜ lim w ⎞⎟
⎝ w→ –2 ⎠
⎝ w→ –2 ⎠

x → –3

=

lim (–3w3 + 7 w2 )

3

4, 5

2

9

w→ –2

w→ –2

x → –3

x → –3

–3w3 + 7 w2

= –3 lim w3 + 7 lim w2

3

=

lim

w→ –2

4, 3

3
⎡ ⎛
⎤
⎞
⎢ 4 ⎜ lim y ⎟ + 8 lim y ⎥
y →2 ⎥
⎢ y →2 ⎠
=⎢ ⎝
⎥
lim y + 4
⎢

⎥
y →2
⎢
⎥
⎣
⎦

2

1/ 3

⎡ 4(2)3 + 8(2) ⎤
=⎢
⎥
2+4
⎣⎢
⎦⎥

=2

8

2

= 5 ⎛⎜ lim x ⎞⎟ + 2 lim x
x → –3
⎝ x→ –3 ⎠

2

= 5(–3)2 + 2(–3) = 39

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or by any means, without permission in writing from the publisher.


12.

lim (2 w 4 – 9 w 3 +19)–1 /2
1

= lim

w→ 5

7

4

2w − 9 w3 + 19
lim 1
w→5
4

=

1, 9

3

2w – 9 w + 19

lim

w→ 5

x→2

19. lim

lim (2w – 9 w3 + 19)

w→ 5

1
lim 2 w4 − lim 9 w3 + lim 19

w→5

w→5

4

2 lim w − 9 lim w3 + 19

w→5

1
4

=

3

1
144

x→2

=

x2 − 4
2

x +4

=

(
lim ( x

2

) 4−4
=

=0
+ 4) 4 + 4

lim x − 4

x→2

2

x→2

= lim ( x − 3) = −1
x→2

( x − 3)( x + 1)
x2 − 2 x − 3
= lim
x →−1
x →−1
x +1
( x + 1)
lim

= lim ( x − 3) = −4

16.

17.

lim

x →−1

x2 + x
x2 + 1

(
lim ( x

2

=

) 0
= =0
+ 1) 2

lim x + x

x →−1

x →−1

2

( x − 1)( x − 2)( x − 3)
x−3
= lim
x →−1 ( x − 1)( x − 2)( x + 7)
x →−1 x + 7

lim

=

−1 − 3
2
=−
−1 + 7
3

2

2

u –u– 6
u– x x+2
= lim
=
5
u → –2 u – 3
u →–2

( u + 2 )( u – x )
u→ –2 ( u + 2)(u – 3)

= lim

x 2 + ux – x – u
( x – 1)( x + u)
= lim

2
x→1 x + 2 x – 3
x →1 ( x – 1)( x + 3)
x + u 1+ u u + 1
= lim
=
=
4
x→1 x + 3 1+ 3

2 x2 – 6 xπ + 4 π2

2( x – π)( x – 2 π)
x→ π
x –π
x→ π ( x – π)( x + π)
2( x – 2π) 2(π – 2 π)
= lim
=
= –1
π+π
x→ π x + π

23. lim

24.

2

lim

2

= lim

(w + 2)(w 2 – w – 6)

w 2 + 4w + 4
( w + 2) 2 ( w – 3)
= lim
= lim ( w – 3)
( w + 2 )2
w→ –2
w→ –2
= –2 – 3 = –5
w→ –2

25. lim

x→a

f 2 ( x) + g 2 ( x)

lim f 2 ( x) + lim g 2 ( x)

x→a

x→a

2

= ⎛⎜ lim f ( x) ⎞⎟ + ⎛⎜ lim g ( x) ⎞⎟
⎝ x →a
⎠ ⎝ x→a
⎠

2

= (3) 2 + (–1)2 = 10
[2 f ( x) – 3 g ( x)]
2 f ( x) – 3g ( x ) xlim
= →a
x → a f ( x) + g ( x)
lim [ f ( x) + g ( x)]

26. lim

x→a

2 lim f ( x) – 3 lim g ( x)
x→a

x→a

lim f ( x) + lim g ( x)

x→a

Section 1.3

= lim

u2 – ux + 2u – 2 x

lim

=

74

( x + 2)( x − 1)
( x + 1)( x − 1)

( x + 3)( x – 17)
x→ –3 x – 4 x – 21
x→ –3 ( x + 3)( x – 7)
x – 17 –3 – 17
= lim
=
=2
–3 – 7
x→ –3 x – 7

=

x →−1

x →1

22. lim

1
12

( x − 3)( x − 2 )
x2 − 5x + 6
14. lim
= lim
x→2
x→2
x−2
( x − 2)

15.

2

2(5)4 − 9(5)3 + 19

lim

13.

21.

2 ⎛⎜ lim w ⎞⎟ − 9 ⎛⎜ lim w ⎞⎟ + 19
⎝ w→ 5 ⎠
⎝ w→5 ⎠
1

= lim

x 2 – 14 x – 51

lim

8

w→5

=

=

1,3

20.

w→5

1

=

2

x −1
x + 2 1+ 2 3
= lim
=

=
x →1 x + 1
1+1 2

4,5

4

=

x2 + x − 2

x →1

1

=

x 2 + 7 x + 10
( x + 2)( x + 5)
= lim
x→2
x
→
2
x+2
x+2
= lim( x + 5) = 7

18. lim

w→ 5

x→a

=

2(3) – 3(–1) 9
=
3 + (–1)
2

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or by any means, without permission in writing from the publisher.


27. lim 3 g ( x) [ f ( x) + 3] = lim 3 g ( x) ⋅ lim [ f ( x) + 3]
x→a

x→a

x→a

= 3 lim g ( x) ⋅ ⎡ lim f ( x) + lim 3⎤ = 3 – 1 ⋅ (3 + 3)
⎢⎣ x → a
x→a
x→a ⎥
⎦

= –6

28. lim [ f ( x) – 3]4 = ⎡⎢ lim ( f ( x) – 3) ⎤⎥
x→a
⎣ x→a
⎦
4

= lim f (t ) + 3 lim g (t )
t →a

⎛
⎞
30. lim [ f (u) + 3g(u)] = ⎜ lim [ f (u) + 3g(u)]⎟
⎝
⎠
u →a
u →a
3

3

3

⎡
⎤
= ⎢ lim f (u ) + 3 lim g(u) ⎥ = [3 + 3( –1)]3 = 0
⎣u→ a
⎦
u →a

3x 2

– 12
3( x – 2 )(x + 2)
31. lim
= lim
x
–
2
x –2
x→2
x→2
= 3 lim (x + 2) = 3(2 + 2) = 12
x→2

(3x 2 + 2 x + 1) – 17
3x 2 + 2 x – 16
= lim
x–2
x–2
x→2
x →2
(3 x + 8)( x – 2)
= lim
= lim (3 x + 8)
x–2
x→2
x →2
= 3 lim x + 8 = 3(2) + 8 = 14
x→2

1
2

= lim

2– x
2x

= lim

–

34.

–

3
4

x–2

3( 4 – x 2 )

= lim

4x2

–3( x + 2 )( x – 2 )

= lim

4x2

x→2
x–2
x–2
–3 ⎛⎜ lim x + 2 ⎞⎟
–3( x + 2)
⎠ = –3(2 + 2)
= lim
= ⎝ x →2
2
2
x→2
4x
4(2)2
4 ⎛⎜ lim x ⎞⎟
⎝ x→2 ⎠
3
=–
4
x→2

x→c

x→c

exist δ 2 and δ 3 such that 0 < x – c < δ 2 ⇒

g ( x) – M <

ε

and 0 < x – c < δ 3 ⇒

L + M +1

ε
L + M +1

. Let

δ = min{δ1 , δ 2 , δ 3 }, then 0 < x – c < δ ⇒
f ( x) g ( x) – LM ≤ g ( x) f ( x) – L + L g ( x ) – M

< ( M + 1)

ε
L + M +1

+L

ε
L + M +1

=ε

Hence,
lim f ( x) g ( x) = LM = ⎛⎜ lim f ( x) ⎞⎟ ⎛⎜ lim g ( x) ⎞⎟

⎝ x →c
⎠ ⎝ x →c
⎠

x→2

36. Say lim g ( x ) = M , M ≠ 0 , and choose
x →c

1
M
.
2
There is some δ1 > 0 such that

ε1 =

1
M or
2

1
1
M < g ( x) < M + M .
2
2
1
1
1
1

M − M ≥ M and M + M ≥ M
2
2
2
2
1
2
1
so g ( x) > M and
<
g ( x)
M
2
M−

x →2

lim

0 < x – c < δ1 , g ( x) < M + 1. Choose ε > 0.
Since lim f (x) = L and lim g(x) = M, there

0 < x − c < δ1 ⇒ g ( x) − M < ε1 =

x–2
2x

x – 2 x→2 x – 2 x →2 x – 2
1
–1

–1
1
= lim –
=
=
=–
2 lim x 2(2)
4
x→2 2 x
x→2

3
x2

x →c

x →c

32. lim

–

f ( x) g ( x) – LM ≤ g ( x) f ( x) – L + L g ( x ) – M

as shown in the text. Choose ε 1 = 1. Since
lim g ( x) = M , there is some δ1 > 0 such that if

f ( x) – L <

= 3 + 3 –1 = 6

33. lim

x→c

M – 1 ≤ M + 1 and M + 1 ≤ M + 1 so for

29. lim ⎡⎣ f (t ) + 3g (t ) ⎤⎦ = lim f (t ) + 3 lim g (t )
t →a
t →a
t →a

1
x

x→c

0 < x – c < δ1 , g ( x) – M < ε1 = 1 or
M – 1 < g(x) < M + 1

4

= ⎡⎢ lim f ( x) – lim 3⎤⎥ = (3 – 3) 4 = 0
x →a ⎦
⎣ x→a

t →a

35. Suppose lim f (x) = L and lim g(x) = M.

Choose ε > 0.
Since lim g(x) = M there is δ 2 > 0 such that
x→c

0 < x − c < δ 2 ⇒ g ( x) − M <

Let δ = min{δ1 , δ 2}, then
0< x–c <δ ⇒

=

1 2
M .
2

1
1
M – g ( x)
–
=
g ( x) M
g ( x) M

1
2
2 1 2
⋅ M ε
g ( x) − M <
g ( x) − M =
2

M g ( x)
M
M2 2

=ε

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or by any means, without permission in writing from the publisher.


Thus, lim
x→c

1
1
1
=
=
.
g(x) M lim g (x)

43.

x→c

Using statement 6 and the above result,
f ( x)
1
lim
= lim f ( x) ⋅ lim
x →c g ( x )
x →c
x →c g ( x )
lim f ( x )
1
.
= lim f ( x) ⋅
= x →c
lim g ( x ) lim g ( x)
x →c
x →c

x→c

x →3+

x→ c

⇔ lim f (x) – lim L = 0
⇔ lim [ f (x) – L] = 0

45.

x→c

2

⎡
⎤
38. lim f (x) = 0 ⇔ ⎢ lim f (x) ⎥ = 0
⎣
⎦
x→c
x→c

x2 – 9

x →3+

46.

2

⇔ lim f ( x) = 0

x2 – 9
x+3

32 – 9
=0
3+3

=

lim

x →1–

x →c

x→c

( x – 3) x 2 – 9

( x – 3) x 2 – 9
= lim
x →3+ ( x – 3)( x + 3)
x →3+

44.

x→ c

x2 – 9

= lim

= lim

x →c

37. lim f (x) = L ⇔ lim f ( x) = lim L

x–3

lim

lim

x → 2+

1+ x
1+1
2
=
=
4 + 4 x 4 + 4(1)
8
( x 2 + 1) x
(3 x − 1)

=

2

(22 + 1) 2
(3 ⋅ 2 − 1)

2

lim ( x − x ) = lim x − lim

x →3−

x →3−

x →3−

=

5⋅ 2
5

2

=

2
5

x = 3− 2 =1

x→c

lim f 2 ( x) = 0

⇔

47.

x →c

⇔ lim

x →c

f 2 ( x) = 0

48.

⇔ lim f ( x) = 0

lim

x
= –1
x

lim

x 2 + 2 x = 32 + 2 ⋅ 3 = 15

x →0 –

x →3+

x→c

2

39. lim x = ⎛⎜ lim x ⎞⎟ =
x →c
⎝ x →c ⎠

lim x

x →c

2

=

lim x 2

x →c

2

x +1
x–5
, g ( x) =
and c = 2, then
x–2
x–2
lim [ f (x) + g (x)] exists, but neither

x→c

lim f (x) nor lim g(x) exists.
x→c

2
, g ( x) = x, and c = 0, then
x

lim [ f (x) ⋅ g( x)] exists, but lim f (x) does

b. If f ( x) =
x→c

x→c

not exist.
41.

42.

76

lim

x → –3+

3+ x
3–3
=
=0
x
–3

π3 + x3
=
x
x → – π+
lim

Section 1.3

1
f ( x)

lim g ( x) = 0 ⇔ lim

1
=0
f ( x)

⇔

If f ( x) =

x→c

f ( x) g ( x) = 1; g ( x) =

x →a

= ⎛⎜ lim x ⎞⎟ = c 2 = c
⎝ x →c ⎠

40. a.

49.

π3 + (– π)3

=0
–π

x →a

1
=0
lim f ( x)

x→a

No value satisfies this equation, so lim f ( x)
x→ a

must not exist.
1⎞
⎛ x
50. R has the vertices ⎜ ± , ± ⎟
⎝ 2
2⎠

Each side of Q has length

x 2 + 1 so the

perimeter of Q is 4 x 2 + 1. R has two sides of
length 1 and two sides of length

x 2 so the

perimeter of R is 2 + 2 x 2 .
lim

x →0 +

=

perimeter of R
2 x2 + 2
= lim
perimeter of Q x →0+ 4 x 2 + 1

2 02 + 2
2

4 0 +1

=

2 1
=
4 2

Instructor’s Resource Manual

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or by any means, without permission in writing from the publisher.


NO = (0 – 0)2 + (1 – 0)2 = 1

51. a.

4. lim

OP = ( x – 0)2 + ( y – 0) 2 = x 2 + y 2
= x2 + x
NP = ( x – 0)2 + ( y – 1)2 = x 2 + y 2 – 2 y + 1
= x + x − 2 x +1
MO = (1 – 0) 2 + (0 – 0) 2 = 1
y2 + x2 – 2 x + 1

= x2 − x + 1
perimeter of ΔNOP
lim
x →0+ perimeter of ΔMOP
= lim
=

1+ 1
1+ 1

1 + x2 + x + x2 – x + 1
=1

x
1
(1)( x) =
2
2

1
x
Area of ΔMOP = (1)( y ) =
2
2

b. Area of ΔNOP =

x

area of ΔNOP
x
= lim 2 = lim
x
+
+
+
MOP
area
of
Δ
x
x →0
x →0
x →0
lim

2

= lim

x →0+

x =0

1.4 Concepts Review
1. 0

0
=0
1

5. lim

sin x 1
sin x 1
1
= lim
= ⋅1 =
2x
2 x →0 x
2
2

sin 3θ
3 sin 3θ 3
sin 3θ
= lim ⋅
= lim
3θ

2 θ →0 3θ
θ →0 2θ
θ →0 2
3
3
= ⋅1 =
2
2

6. lim

sin 3θ
sin 3θ
cos θ sin 3θ
= lim sin θ = lim
θ → 0 tan θ
θ →0
θ →0
sin θ
cos θ
⎡
sin 3θ 1 ⎤
= lim ⎢cos θ ⋅ 3 ⋅
⋅ sin θ ⎥
θ →0⎢
3θ
θ ⎥
⎣
⎦
⎡

sin 3θ 1
= 3 lim ⎢cos θ ⋅
⋅ sin θ
θ →0 ⎢
3θ
θ
⎣

⎤
⎥ = 3 ⋅1 ⋅1 ⋅1 = 3
⎥⎦

sin 5θ

sin 5θ
tan 5θ
= lim cos 5θ = lim
θ → 0 sin 2θ
θ → 0 sin 2θ
θ → 0 cos 5θ sin 2θ
sin 5θ 1 2θ ⎤
⎡ 1
= lim ⎢
⋅5⋅
⋅ ⋅
5θ 2 sin 2θ ⎥⎦
θ →0 ⎣ cos 5θ
5
sin 5θ 2θ ⎤
⎡ 1

= lim ⎢
⋅
⋅
2 θ →0 ⎣ cos 5θ 5θ sin 2θ ⎥⎦
5
5
= ⋅1⋅1⋅1 =
2
2

8. lim

cot πθ sin θ
= lim
θ →0
θ →0
2 sec θ

9. lim

cos πθ
sin πθ

sin θ

2
cos θ

cos πθ sin θ cos θ
2sin πθ

⎡ cos πθ cos θ sin θ 1 πθ ⎤
= lim ⎢
⋅
⋅ ⋅
θ π sin πθ ⎥⎦
2
θ →0 ⎣
1
sin θ πθ ⎤
⎡
=
⋅
lim ⎢cos πθ cos θ ⋅
θ sin πθ ⎥⎦
2 π θ →0 ⎣
1
1
⋅1⋅1⋅1⋅1 =
=
2π
2π
= lim

2. 1

θ →0

3. the denominator is 0 when t = 0 .
4. 1

Problem Set 1.4
cos x 1
= =1
x →0 x + 1 1

sin 2 3t
9t sin 3t sin 3t
= lim ⋅
⋅
= 0 ⋅1 ⋅1 = 0
t →0
t
→
0
2t
2 3t
3t

1. lim

2.

3 x tan x
3x (sin x / cos x)
3x
= lim
= lim
x →0
x → 0 cos x
sin x

sin x

7. lim

1 + x2 + x + x2 + x – 2 x + 1

x → 0+

=

x →0

2

MP = ( x – 1)2 + ( y – 0) 2 =

x →0

lim θ cosθ =

θ →π / 2

π
2

10. lim

⋅0 = 0

cos 2 t

cos 2 0
1
=
=
=1
t →0 1 + sin t 1 + sin 0 1 + 0

3. lim

tan 2 3t
sin 2 3t
= lim
t →0
t →0 (2t )(cos 2 3t )
2t

11. lim

= lim
t →0

3(sin 3t ) sin 3t
⋅
= 0 ⋅1 = 0
3t

2 cos 2 3t

tan 2t
0

=
=0
t → 0 sin 2t − 1
−1

12. lim

Instructor’s Resource Manual

Section 1.4

77

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form
or by any means, without permission in writing from the publisher.


sin(3t ) + 4t
4t ⎞
⎛ sin 3t
= lim ⎜
+
⎟
t →0 ⎝ t sec t
t sec t
t sec t ⎠
sin 3t
4t
= lim
+ lim

t →0 t sec t t →0 t sec t
sin 3t
= lim 3cos t ⋅
+ lim 4 cos t
t →0
t →0
3t
= 3 ⋅1 + 4 = 7

13. lim

t →0

14.

sin 2 θ
sin θ sin θ
lim
= lim
2
θ
θ →0 θ
θ →0 θ
sin θ
sin θ
= lim
× lim
= 1× 1 = 1
θ
θ →0

θ →0 θ

19. lim 1 +
x →0

sin x
=2
x

20. The result that lim cos t = 1 was established in
t →0

the proof of the theorem. Then
lim cos t = lim cos(c + h)
t →c

h →0

= lim (cos c cos h − sin c sin h)
h →0

= lim cos c lim cos h − sin c lim sin h

15. lim x sin (1/ x ) = 0

h →0

x →0

h →0

h→0

= cos c
lim sin t
sin t t →c
sin c
=
=
= tan c
t → c cos t
lim cos t cos c

21. lim tan t = lim
t →c

t →c

lim cot t = lim
t →c

(

)

16. lim x sin 1/ x 2 = 0
x →0

t →c

lim cos t

cos t t →c
cos c
=
=
= cot c
sin t lim sin t sin c
t →c

1
1
=
= sec c
cos t cos c
1
1
lim csc t = lim
=
= csc c
t →c
t →c sin t
sin c

22. lim sec t = lim
t →c

t →c

23. BP = sin t , OB = cos t

area( ΔOBP) ≤ area (sector OAP)
≤ area (ΔOBP) + area( ABPQ)

(

)

17. lim 1 − cos 2 x / x = 0
x →0

1
1
1
OB ⋅ BP ≤ t (1) 2 ≤ OB ⋅ BP + (1 – OB ) BP
2
2
2
1
1
1
sin t cos t ≤ t ≤ sin t cos t + (1 – cos t ) sin t
2
2
2
t
≤ 2 – cos t
sin t
1
sin t
1

π
π
≤
≤
for − < t < .
2 – cos t
t
cos t
2
2
1
sin t
1
lim
≤ lim
≤ lim
t →0 2 – cos t t →0 t
t →0 cos t
sin t
≤1
1 ≤ lim
t →0 t
sin t
= 1.
Thus, lim
t →0 t

cos t ≤

18. lim cos 2 x = 1

x →0

78

Section 1.4

Instructor’s Resource Manual

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or by any means, without permission in writing from the publisher.


24. a.

Written response
6.

b.

c.

1
1
AB ⋅ BP = (1 − cos t ) sin t
2
2
sin t (1 − cos t )
=
2
1

1
t sin t cos t
E = t (1)2 – OB ⋅ BP = –
2
2
2
2
D sin t (1 – cos t )
=
E
t – sin t cos t
D=

⎛D⎞
lim ⎜ ⎟ = 0.75
+
t →0 ⎝ E ⎠

3

2 x – 100 x

x →∞

8.

10.

1. x increases without bound; f(x) gets close to L as
x increases without bound

4. x = 6; vertical

sin 2 θ

lim

θ →∞ θ 2 – 5

1.

2.

3.

4.

5.

lim

x →∞

x
1
= lim
=1
x – 5 x →∞ 1 – 5

lim

x

x2

x →∞ 5 – x3

lim

t2

t →–∞ 7 − t

lim

t →–∞ t

2

= lim

x →∞ 5
x3

x →∞

= lim

θ →∞ 1 – 5
θ2

x

2

= lim

= 0 so lim

sin 2 θ

θ →∞ θ 2 – 5

=0

3 x3 / 2 + 3 x
2 x3 / 2

x →∞

3

=

2

πx3 + 3x
πx3 + 3 x
lim 3
= 3 lim

x →∞
x →∞ 2 x3 + 7 x
2 x3 + 7 x
3
x2
+ 72
x

π+
2

=3

π
2

=0
2

13.
1

t →–∞ 7
t2

θ2

= lim

2 x3

3+ 3

x →∞

–1

=π

; 0 ≤ sin 2 θ ≤ 1 for all θ and

3 x3 + 3 x

lim

= 3 lim

1
x

π

θ →–∞ 1 – 5
θ

1

1

x →∞

12.

x →∞

3– 1 3
3 x3 – x 2
x =
= lim
lim
3
2
x → ∞ πx – 5 x
x→∞π– 5 π
x

= lim

Problem Set 1.5

1
1
=
2 – 100
2
x

= lim

= lim

θ → – ∞ θ 5 – 5θ 4

θ →∞ θ 2 – 5

11.

2

πθ 5

lim

lim

2. f(x) increases without bound as x approaches c
from the right; f(x) decreases without bound as x
approaches c from the left
3. y = 6; horizontal

x3

7. lim

9.

1.5 Concepts Review

1
x2
lim

= lim
=1
2
8
x → ∞ x – 8 x + 15 x → ∞ 1 – + 15
x x2

−1

= −1

t
1
= lim
=1
– 5 t →–∞ 1 – 5
t

= 3 lim

14.

x2
x2
= lim
x →∞ ( x – 5)(3 – x) x →∞ − x 2 + 8 x − 15

Instructor’s Resource Manual

1

x2

+8

x →∞ 1 + 4
x2

lim

1
= lim
= –1
8
x → ∞ −1 + − 15
x x2

2

1 + 8x
1 + 8x
lim 3
= 3 lim
2
x →∞ x + 4
x →∞ x 2 + 4

lim

x →∞

=

= 38 =2

x2 + x + 3
=
( x –1)( x + 1)

lim

x →∞

1 + 1x + 32
x

1 – 12

lim

x2 + x + 3

x →∞

x 2 –1

= 1 =1

x

15.

lim

n →∞

n
1
1
= lim
=
2n + 1 n→∞ 2 + 1 2
n

Section 1.5

79

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or by any means, without permission in writing from the publisher.


16. lim

n →∞

n2
n2 + 1

1

= lim

n →∞

1+

=

1
n2

1
=1
1+ 0

23.

lim n
n2
n
∞
n →∞
= lim
=
=
=∞
1
n →∞ n + 1 n →∞
1 ⎞ 1+ 0
⎛

1+
lim 1 +
n n →∞ ⎜⎝ n ⎟⎠

17. lim

18. lim

n →∞

n
n2 + 1

n →∞

1+

lim

=

1
n2

x →∞

x →∞

0
=0

1+ 0

x →∞

x2 + 3

= lim

2 + 1x

x →∞

x 2 +3
x

=

20.

21.

2
1

x →∞

1+

2

3
x2

lim

x →∞

x →∞

= lim

x →∞

2 x 2 + 3 – (2 x 2 – 5)
2 x2 + 3 + 2 x2 – 5
8
2

2 x + 3 + 2 x2 − 5
8
x

2+

3
x2

x →∞

8

x
2 x 2 +3 + 2 x 2 –5
x2

+ 2–

22.
lim ⎛⎜ x 2 + 2 x − x ⎞⎟
⎠

x →∞ ⎝

⎛ x 2 + 2 x – x ⎞⎛ x 2 + 2 x + x ⎞
⎜
⎟⎜
⎟
⎠⎝
⎠
= lim ⎝
2
x →∞
x + 2x + x

= lim

x →∞

= lim

x →∞

2

x + 2x – x

2

= lim

2x

x 2 + 2 x + x x→∞ x 2 + 2 x + x
2
2
= =1
1+ 2 +1 2

n →∞

+

bn
xn

1
1
1+ 2
n

=

=

a0
b0
1

1+ 0

=1

t → –3+

29. As t → 3– , t 2 → 9 while 9 – t 2 → 0+.
t →3–

t2

9 – t2

=∞
+

30. As x → 3 5 , x 2 → 52 / 3 while 5 – x3 → 0 – .
x→3 5

5
x2

= lim

an
xn

t2 – 9
(t + 3)(t – 3)
= lim
+
+
t +3
t → –3 t + 3
t → –3
= lim (t – 3) = –6

lim

=0

+ …+

bn –1
x n –1

+

lim

lim

= lim

an –1
x n –1

27. As x → 4+ , x → 4 while x – 4 → 0 .
x
lim
=∞
+
x
–
4
x →4
28.

⎛ 2 x 2 + 3 – 2 x 2 – 5 ⎞⎛ 2 x 2 + 3 + 2 x 2 – 5 ⎞
⎜
⎟⎜
⎟
⎠⎝
⎠
= lim ⎝
2
2
x →∞
2x + 3 + 2x – 5

= lim

+ …+

+

+ 12

lim ⎛⎜ 2 x 2 + 3 – 2 x 2 – 5 ⎞⎟
⎠
x →∞ ⎝

x →∞

= –∞

n2
∞
n
n3/ 2
= lim
= =∞
3
n →∞
1
2
1
n + 2n + 1
1+ 2 + 3
n
n

26. lim

2x +1
x
= lim
= lim
=0
x+4
x →∞ 1 + 4
x →∞ 1 + 4
x
x

= lim

b0 +

b1
x

n +1

n →∞

2
x

y→–∞ 1 –

2

=2
2 x +1
x2

a1
x

2

2 + 1x

= lim

a0 +

n

25. lim

19. For x > 0, x = x 2 .
2x + 1

= lim

b0 x n + b1 x n –1 +…+ bn –1 x + bn

n →∞

lim

y2 – 2 y + 2

a0 x n + a1 x n –1 +…+ an –1 x + an

= lim

1
n

= lim

24.

lim

y→– ∞

1
y2
2+ 2
y y2

9y +

9 y3 + 1

x2
+

5 – x3

= –∞

31. As x → 5– , x 2 → 25, x – 5 → 0 – , and
3 – x → –2.
x2
lim
=∞
x →5 – ( x – 5)(3 – x)
32. As θ → π+ , θ 2 → π2 while sin θ → 0− .

θ2
= −∞
θ →π+ sin θ
lim

x

80

Section 1.5

Instructor’s Resource Manual

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or by any means, without permission in writing from the publisher.


−
33. As x → 3− , x3 → 27, while x − 3 → 0 .

43.

3

lim

x →3−

x
= −∞
x−3

π+
π2
, πθ →
while cos θ → 0 – .
2
2
πθ
= –∞
lim
π + cos θ

34. As θ →

θ→

35.

3
3
= 0, lim
= 0;
x +1
x→ – ∞ x + 1
Horizontal asymptote y = 0.
3
3
lim
= ∞, lim
= – ∞;
x → –1+ x + 1
x → –1– x + 1
Vertical asymptote x = –1
lim

x →∞

2

lim

x →3–

x2 – x – 6
( x + 2)( x – 3)
= lim
x–3
x–3

x →3–

= lim ( x + 2) = 5
x →3 –

36.

x2 + 2 x – 8

lim

2

x → 2+

= lim

x → 2+

x –4
x+4 6 3
= lim
= =
x → 2+ x + 2 4 2

( x + 4)( x – 2)
( x + 2)( x – 2)

37. For 0 ≤ x < 1 , x = 0 , so for 0 < x < 1,

thus lim

x →0 +

x
x

44.

x
x

lim

3

x →∞ ( x + 1)

2

3

= 0, lim

x → – ∞ ( x + 1) 2

= 0;

Horizontal asymptote y = 0.
3

3
lim
= ∞, lim
= ∞;
2
2
–
+
x → –1 ( x + 1)
x → –1 ( x + 1)
Vertical asymptote x = –1

=0

=0

38. For −1 ≤ x < 0 , x = −1 , so for –1 < x < 0,
x
1
thus lim
= ∞.
−
x
x
x
x →0
1
(Since x < 0, – > 0. )
x
x

=−

39. For x < 0, x = – x, thus
lim

x →0 –

x
x

= lim

x →0 –

–x
= –1
x

40. For x > 0, x = x, thus lim

x →0 +

45.

x
x

= lim

x →0 +

x
=1
x

41. As x → 0 – , 1 + cos x → 2 while sin x → 0 – .
1 + cos x
lim
= –∞
–
sin x
x →0

lim

x →∞

2x
2
= lim
= 2,
x – 3 x→∞ 1 – 3
x

2x
2
lim
= lim
= 2,

x →−∞ x – 3 x →−∞ 1 – 3
x

Horizontal asymptote y = 2
2x
2x
lim
= ∞, lim
= – ∞;
x →3+ x – 3
x →3– x – 3
Vertical asymptote x = 3

42. –1 ≤ sin x ≤ 1 for all x, and
1
sin x
lim = 0, so lim
= 0.
x →∞ x
x →∞ x

Instructor’s Resource Manual

Section 1.5

81

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or by any means, without permission in writing from the publisher.

46.

lim

3
2

x →∞ 9 –

3

= 0, lim

2

49. f ( x ) = 2 x + 3 –

= 0;

x→ – ∞ 9 – x
x
Horizontal asymptote y = 0
3
3
lim
= – ∞, lim
= ∞,
2
2

–
+
x →3 9 – x
x →3 9 – x
3
3
lim
= ∞, lim
= – ∞;
2
2
x → –3+ 9 – x
x → –3– 9 – x
Vertical asymptotes x = –3, x = 3

50.

14

x →∞ 2 x

2

f ( x) = 3x + 4 –

4x + 3
x2 + 1

, thus

We say that lim f ( x) = – ∞ if to each
x →c +

negative number M there corresponds a δ > 0
such that 0 < x – c < δ ⇒ f(x) < M.

14

= 0, lim

, thus

⎡ 4x + 3⎤
lim [ f ( x) – (3 x + 4)] = lim ⎢ –
⎥
x →∞
x →∞ ⎣ x 2 + 1 ⎦
⎡ 4+ 3 ⎤
x x2 ⎥
= lim ⎢ –
=0.
⎢
x →∞
1 + 12 ⎥
⎢⎣
⎥
x ⎦
The oblique asymptote is y = 3x + 4.

= 0;

x→ – ∞ 2 x2 + 7
+7
Horizontal asymptote y = 0
2
Since 2x + 7 > 0 for all x, g(x) has no vertical
asymptotes.
lim

x –1

1 ⎤
⎡
lim [ f ( x) – (2 x + 3)] = lim ⎢ –
⎥=0
3
x →∞
x →∞ ⎣ x –1 ⎦
The oblique asymptote is y = 2x + 3.

51. a.

47.

1
3

b. We say that lim f ( x) = ∞ if to each
x →c –

positive number M there corresponds a δ > 0

such that 0 < c – x < δ ⇒ f(x) > M.
We say that lim f ( x) = ∞ if to each

52. a.

x →∞

positive number M there corresponds an
N > 0 such that N < x ⇒ f(x) > M.
b. We say that lim f ( x ) = ∞ if to each
x → –∞

positive number M there corresponds an
N < 0 such that x < N ⇒ f(x) > M.
53. Let ε > 0 be given. Since lim f ( x ) = A, there is
x →∞

48.

lim

x →∞

lim

x→ – ∞

2x
2

x +5

= lim

2x
x2 + 5

x →∞

= lim

2
1+

x→ – ∞

5
x2

=

2
– 1+

5
x2

2
1
=

a corresponding number M1 such that

= 2,
2
– 1

ε

x > M1 ⇒ f ( x) – A < . Similarly, there is a
2

= –2

Since x 2 + 5 > 0 for all x, g(x) has no vertical
asymptotes.

ε

number M2 such that x > M 2 ⇒ g ( x) – B < .
2
Let M = max{M1 , M 2 } , then
x > M ⇒ f ( x) + g ( x) – ( A + B)
= f ( x) – A + g ( x) – B ≤ f ( x) – A + g ( x) – B

ε

ε

=ε

2 2
Thus, lim [ f ( x) + g ( x)] = A + B
<

+

x →∞

54. Written response

82

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or by any means, without permission in writing from the publisher.


55. a.

lim sin x does not exist as sin x oscillates

x →∞

56.

between –1 and 1 as x increases.
1

, then as x → ∞, u → 0+.
x
1
lim sin = lim sin u = 0
x u →0 +
x →∞

b. Let u =

c.

1
Let u = , then as x → ∞, u → 0+.
x
1
1
sin u
lim x sin = lim sin u = lim
=1
+
x u → 0+ u
u
x →∞
u →0

d. Let u =
lim x

3/ 2

x →∞

e.

h.

59.

60.
3/ 2

1 − v2 / c2

v →c

3x 2 + x +1 3
=
2
2
x →∞ 2x –1
2 x 2 – 3x

lim

=

2

5x + 1

x→ – ∞

2
5

3
lim ⎛⎜ 2 x 2 + 3x – 2 x 2 – 5 ⎞⎟ = –
⎠
2 2

x→ – ∞ ⎝

2x +1

lim

x →∞

3x 2 + 1

sin u

2

=

3

10

⎡⎛ 1 ⎞⎛ sin u ⎞⎤
⎟⎟⎜
= lim+ ⎢⎜⎜
⎟⎥ = ∞
u →0 ⎣
⎢⎝ u ⎠⎝ u ⎠⎦⎥

As x → ∞, sin x oscillates between –1 and 1,
1
while x –1/ 2 =
→ 0.
x

62.

⎛ 1⎞
lim ⎜1 + ⎟ = e ≈ 2.718
x⎠
x →∞ ⎝

–1/ 2

sin x = 0

1
, then
x
⎛π 1⎞
⎛π
⎞

lim sin ⎜ + ⎟ = lim+ sin ⎜ + u ⎟
x→∞
x
6
6
u
→
0
⎝
⎠
⎝
⎠
π 1
= sin =
6 2

1
1⎞
⎛
→ ∞, so lim sin ⎜ x + ⎟
x
x⎠
x →∞
⎝
does not exist. (See part a.)
1⎞
1
1
⎛
sin ⎜ x + ⎟ = sin x cos + cos x sin

x⎠
x
x
⎝
⎡ ⎛
1⎞
⎤
lim ⎢sin ⎜ x + ⎟ – sin x ⎥
x
x →∞ ⎣
⎝
⎠
⎦

⎡
1 ⎞
1⎤
⎛
= lim ⎢sin x ⎜ cos –1⎟ + cos x sin ⎥
x ⎠
x⎦
x →∞ ⎣
⎝
1
1
As x → ∞, cos → 1 so cos –1 → 0.
x
x
1
From part b., lim sin = 0.

x
x →∞
As x → ∞ both sin x and cos x oscillate
between –1 and 1.
⎡ ⎛
1⎞
⎤
lim ⎢sin ⎜ x + ⎟ – sin x ⎥ = 0.
x⎠
x →∞ ⎣
⎝
⎦

Instructor’s Resource Manual

=1
x

⎛ 1⎞
63. lim ⎜ 1 + ⎟
x⎠
x →∞ ⎝

Let u =

As x → ∞, x +

=∞

lim

⎛ 1⎞
lim ⎜ 1 + ⎟
x⎠
x →∞ ⎝

lim x

g.

1
⎛1⎞
sin = lim+ ⎜ ⎟
x u →0 ⎝ u ⎠

58.

v →c

61.

x →∞

f.

1
, then
x

57.

m0

lim− m(v) = lim−

64.

65.

66.

67.

68.

⎛ 1⎞
lim ⎜1 + ⎟
x⎠
x →∞ ⎝

70.

71.

=∞
sin x

=1

sin x – 3

lim

x →3–

= –1

x–3

sin x – 3

lim

x →3–

tan( x – 3)

= –1

lim

cos( x – 3)
= –∞
x–3

lim

cos x
= –1
x – π2

x →3–

x→ π
2

69.

x2

+

lim (1 + x )

x →0 +

1
x

= e ≈ 2.718

lim (1 + x )1/ x = ∞

x → 0+

lim (1 + x ) x = 1

x →0+

Section 1.5

83

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or by any means, without permission in writing from the publisher.


13.

1.6 Concepts Review

t →3+

lim f (t ) = lim (t – 3) = 0

t →3 –

1. lim f ( x)
x →c

t →3–

lim f (t ) = f (3); continuous

t →3

2. every integer
3.

lim f (t ) = lim (3 – t ) = 0

t →3+

lim f ( x) = f (a); lim f ( x) = f (b)

x→a+

14.

x →b –

lim f (t ) = lim (3 – t )2 = 0

t →3+

t →3+

lim f (t ) = lim (t 2 – 9) = 0

t →3–

4. a; b; f(c) = W

t →3 –

lim f (t ) = f (3); continuous

t →3

15. lim f ( x) = −2 = f (3); continuous

Problem Set 1.6

t →3

1. lim[( x – 3)( x – 4)] = 0 = f (3); continuous
x →3

2. lim ( x 2 – 9) = 0 = g (3); continuous
x →3

3

3. lim

x →3 x – 3

and h(3) do not exist, so h(x) is not

continuous at 3.

16. g is discontinuous at x = –3, 4, 6, 8; g is left
continuous at x = 4, 8; g is right continuous at
x = –3, 6
17. h is continuous on the intervals
(−∞, −5), [ −5, 4] , (4, 6), [ 6,8] , (8, ∞)
x 2 – 49
( x – 7)( x + 7)
= lim
= lim ( x + 7)

x–7
x →7 x – 7
x →7
x →7
= 7 + 7 = 14
Define f(7) = 14.

18. lim

4. lim t – 4 and g(3) do not exist, so g(t) is not
t →3

continuous at 3.
t –3

and h(3) do not exist, so h(t) is not
t –3
continuous at 3.

5. lim

2 x 2 –18
2( x + 3)( x – 3)
= lim
3– x
x →3 3 – x
x →3
= lim[–2( x + 3)] = –2(3 + 3) = –12

19. lim

t →3

x →3

Define f(3) = –12.

6. h(3) does not exist, so h(t) is not continuous at 3.
7. lim t = 3 = f (3); continuous
t →3

20. lim

t →3

21. lim

t →1

t 3 – 27
(t – 3)(t 2 + 3t + 9)
= lim
t –3
t →3 t – 3
t →3
= lim(t 2 + 3t + 9) = (3)2 + 3(3) + 9 = 27 = r (3)

22.

12. From Problem 11, lim r (t ) = 27, so r(t) is not

t →3

continuous at 3 because lim r (t ) ≠ r (3).
t →3

t –1

–1)( t + 1)
1
Define H(1) = .
2
t →1 (t

11. lim

continuous

t –1
( t –1)( t + 1)
= lim
t –1 t →1 (t –1)( t + 1)

= lim

10. f(3) does not exist, so f(x) is not continuous at 3.

t →3

=1

θ
Define g(0) = 1
θ →0

8. lim t – 2 = 1 = g (3); continuous
9. h(3) does not exist, so h(t) is not continuous at 3.

sin(θ )

= lim

t →1

1
t +1

=

1
2

x4 + 2 x2 – 3
( x 2 –1)( x 2 + 3)
= lim
x +1
x +1
x → –1
x → –1
lim

( x + 1)( x – 1)( x 2 + 3)
x +1
x → –1

= lim

= lim [( x – 1)( x 2 + 3)]
x → –1

= (–1 – 1)[(–1)2 + 3] = –8
Define φ(–1) = –8.

84

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or by any means, without permission in writing from the publisher.


23.

⎛ x2 – 1 ⎞
⎛ ( x – 1)( x + 1) ⎞
lim sin ⎜
⎟ = lim sin ⎜
⎟
⎜

⎟
x +1
x → –1
⎠
⎝ x + 1 ⎠ x→ –1 ⎝
= lim sin( x –1) = sin(–1 – 1) = sin(−2) = – sin 2

37.

x → –1

Define F(–1) = –sin 2.
24. Discontinuous at x = π ,30
25.

33 – x 2
(π – x)( x – 3)
Discontinuous at x = 3, π
f ( x) =

38.

26. Continuous at all points
27. Discontinuous at all θ = nπ + π where n is any
2
integer.
28. Discontinuous at all u ≤ −5

39.

29. Discontinuous at u = –1
30. Continuous at all points
31. G ( x) =

1
(2 – x)(2 + x)

Discontinuous on (−∞, −2] ∪ [2, ∞)
32. Continuous at all points since
lim f ( x) = 0 = f (0) and lim f ( x) = 1 = f (1).
x →0

40.

x →1

33. lim g ( x ) = 0 = g (0)
x →0

lim g ( x) = 1, lim g ( x) = –1

x →1+

x →1–

lim g(x ) does not exist, so g(x) is discontinuous
x→1

at x = 1.
34. Discontinuous at every integer

35. Discontinuous at t = n +

1
where n is any integer
2

Discontinuous at all points except x = 0, because
lim f ( x ) ≠ f (c) for c ≠ 0 . lim f ( x ) exists only
x →c

x →c

at c = 0 and lim f ( x) = 0 = f (0) .
x →0

36.
41. Continuous.
42. Discontinuous: removable, define f (10) = 20
43. Discontinuous: removable, define f (0) = 1
44. Discontinuous: nonremovable.
45. Discontinuous, removable, redefine g (0) = 1
46. Discontinuous: removable, define F (0) = 0

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Section 1.6

85

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or by any means, without permission in writing from the publisher.


47. Discontinuous: nonremovable.
48. Discontinuous: removable, define f (4) = 4
49. The function is continuous on the intervals
( 0,1] , (1, 2], (2,3], …

52. Let f ( x) = x3 + 3 x − 2. f is continuous on [0, 1].
f(0) = –2 < 0 and f(1) = 2 > 0. Thus, there is at
least one number c between 0 and 1 such that
x 3 + 3x − 2 = 0.
53. Because the function is continuous on [ 0,2π ] and
(cos 0)03 + 6sin 5 0 – 3 = –3 < 0,

Cost $

0.60

(cos 2π)(2π)3 + 6sin 5 (2π) – 3 = 8π3 – 3 > 0, there
is at least one number c between 0 and 2π such

0.48

that (cos t )t 3 + 6sin 5 t – 3 = 0.

0.72

0.36

54. Let f ( x ) = x − 7 x + 14 x − 8 . f(x) is
continuous at all values of x.
f(0) = –8, f(5) = 12
Because 0 is between –8 and 12, there is at least
one number c between 0 and 5 such that
3

0.24
0.12
1
3
5
2
4
6
Length of call in minutes

50. The function is continuous on the intervals
[0, 200], (200,300], (300, 400], …

2

f ( x ) = x 3 − 7 x 2 + 14 x − 8 = 0 .

This equation has three solutions (x = 1,2,4)

Cost $
80
60
40

55. Let f ( x ) = x − cos x. . f(x) is continuous at all

20

100 200 300 400 500
Miles Driven

51. The function is continuous on the intervals
(0, 0.25], (0.25, 0.375], (0.375, 0.5], …

values of x ≥ 0.
f(0) = –1, f(π/2) = π / 2
Because 0 is between –1 and π / 2 , there is at
least one number c between 0 and π/2 such that
f ( x ) = x − cos x = 0.
The interval [0.6,0.7] contains the solution.

Cost $
4
3
2
1

0.25

0.5
0.75
Miles Driven

1

56. Let f ( x) = x5 + 4 x3 – 7 x + 14
f(x) is continuous at all values of x.
f(–2) = –36, f(0) = 14
Because 0 is between –36 and 14, there is at least
one number c between –2 and 0 such that
f ( x) = x5 + 4 x3 – 7 x + 14 = 0.

86

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or by any means, without permission in writing from the publisher.


57. Suppose that f is continuous at c, so
lim f ( x) = f (c). Let x = c + t, so t = x – c, then
x →c

as x → c , t → 0 and the statement
lim f ( x) = f (c) becomes lim f (t + c ) = f (c).
x →c

t →0

Suppose that lim f (t + c) = f (c) and let x = t +

t→ 0

c, so t = x – c. Since c is fixed, t → 0 means that
x → c and the statement lim f (t + c) = f (c)
t →0

becomes lim f ( x) = f (c) , so f is continuous at
x →c

c.
58. Since f(x) is continuous at c,
lim f ( x) = f (c) > 0. Choose ε = f ( c ) , then
x →c

there exists a δ > 0 such that
0 < x − c < δ ⇒ f ( x) − f (c) < ε .
Thus, f ( x ) − f ( c ) > −ε = − f ( c ) , or f ( x ) > 0 .
Since also f ( c ) > 0 , f ( x ) > 0 for all x in
(c − δ , c + δ ).

59. Let g(x) = x – f(x). Then,
g(0) = 0 – f(0) = –f(0) ≤ 0 and g(1) = 1 – f(1) ≥ 0
since 0 ≤ f(x) ≤ 1 on [0, 1] . If g(0) = 0, then
f(0) = 0 and c = 0 is a fixed point of f. If g(1) = 0,
then f(1) = 1 and c = 1 is a fixed point of f. If
neither g(0) = 0 nor g(1) = 0, then g(0) < 0 and
g(1) > 0 so there is some c in [0, 1] such that
g(c) = 0. If g(c) = 0 then c – f(c) = 0 or
f(c) = c and c is a fixed point of f.
60. For f(x) to be continuous everywhere,

f(1) = a(1) + b = 2 and f(2) = 6 = a(2) + b
a+b=2
2a + b = 6
– a = –4
a = 4, b = –2

63. Let f(x) be the difference in times on the hiker’s
watch where x is a point on the path, and suppose
x = 0 at the bottom and x = 1 at the top of the
mountain.
So f(x) = (time on watch on the way up) – (time
on watch on the way down).
f(0) = 4 – 11 = –7, f(1) = 12 – 5 = 7. Since time is
continuous, f(x) is continuous, hence there is
some c between 0 and 1 where f(c) = 0. This c is
the point where the hiker’s watch showed the
same time on both days.
⎡ π⎤
64. Let f be the function on ⎢0, 2 ⎥ such that f(θ) is
⎣
⎦
the length of the side of the rectangle which
makes angle θ with the x-axis minus the length of
the sides perpendicular to it. f is continuous on
⎡ π⎤
⎢0, 2 ⎥ . If f(0) = 0 then the region is
⎣
⎦
circumscribed by a square. If f(0) ≠ 0, then
⎛π ⎞

observe that f (0) = − f ⎜ ⎟ . Thus, by the
⎝2⎠
Intermediate Value Theorem, there is an angle

θ 0 between 0 and

π

such that f (θ 0 ) = 0.
2
Hence, D can be circumscribed by a square.
65. Yes, g is continuous at R .
lim g ( r ) =

r →R−

= lim g ( r )

GMm

r →R+

R2

66. No. By the Intermediate Value Theorem, if f
were to change signs on [a,b], then f must be
0 at some c in [a,b]. Therefore, f cannot
change sign.
67. a.

f(x) = f(x + 0) = f(x) + f(0), so f(0) = 0. We
want to prove that lim f (x) = f (c), or,
x→c

equivalently, lim [ f (x) – f (c)] = 0. But
x→c

61. For x in [0, 1], let f(x) indicate where the string
originally at x ends up. Thus f(0) = a, f(1) = b.
f(x) is continuous since the string is unbroken.
Since 0 ≤ a, b ≤ 1 , f(x) satisfies the conditions of
Problem 59, so there is some c in [0, 1] with
f(c) = c, i.e., the point of string originally at c
ends up at c.
62. The Intermediate Value Theorem does not imply
the existence of a number c between –2 and 2
such that f (c ) = 0. The reason is that the

function f ( x ) is not continuous on [ −2, 2] .

Instructor’s Resource Manual

f(x) – f(c) = f(x – c), so
lim[ f ( x) – f (c)] = lim f ( x – c). Let
x →c

x →c

h = x – c then as x → c, h → 0 and
lim f ( x – c) = lim f (h) = f (0) = 0. Hence

x →c

h →0

lim f (x) = f (c) and f is continuous at c.
x→c

Thus, f is continuous everywhere, since c
was arbitrary.
b. By Problem 43 of Section 0.5, f(t) = mt for
all t in Q. Since g(t) = mt is a polynomial
function, it is continuous for all real
numbers. f(t) = g(t) for all t in Q, thus
f(t) = g(t) for all t in R, i.e. f (t ) = mt.

Section 1.6

87

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or by any means, without permission in writing from the publisher.