zill calculo 4e manual de solucionario c02(1)

Chapter 2

Limit of a Function

2.1 Limits — An Informal Approach

1. limx→2

(3x+ 2) = 8

-4 4

4

8

–!

!

2!

2

• (2, 8)

2. limx→2

(x2 − 1) = 3

-4 -2 2 4

3

6

–!

!

2!

2

(–2, 3) •

3. No limit as x → 0.

-4 -2 2 4

-4

4

–!

!

2

!

2 4. limx→5

√x− 1 = 2

-5 5

-5

5

–!

!

2!

2

(5, 2)

•

5. limx→1

x2 − 1

x− 1= lim

x→1(x+ 1) = 2

-5 5

-5

5

–!

!

2!

2

(1, 2)

6. limx→0

x2 − 3x

x= lim

x→0(x− 3) = −3

-5 5

-5

5

–!

!

2!

2

(0, 3)

73

74 CHAPTER 2. LIMIT OF A FUNCTION


3 6

-3

3


-3 3

-3

3

9. limx→0

x3

x= 0

-3 3

-3

3

(0, 0)

10. limx→1

x4 − 1

x2 − 1= 2

-3 3

-3

3

(1, 2)

11. limx→0

f(x) = 3

-3 3

-3

3(0, 3) •


-3 3

-3

3

–!

!

2

!

2

13. limx→2

f(x) = 0

-3 3

-3

3

(2, 0)


-3 3

-3

3

15. (a) 1 (b) -1 (c) 2 (d) doesn’t exist

16. (a) 0 (b) 3 (c) 3 (d) 3

17. (a) 2 (b) -1 (c) -1 (d) -1

18. (a) doesn’t exist (b) 3 (c) -2 (d) doesn’t exist

19. Correct

20. Incorrect; limx→0+

4√x = 0

2.1. LIMITS — AN INFORMAL APPROACH 75

21. Incorrect; limx→1−

√1− x = 0

22. Correct

23. Incorrect; limx→0+

�x� = 0

24. Correct

25. Correct


cos−1 x = 0


√9− x2 = 0

28. Correct

29. (a) Does not exist (b) 0 (c) 3 (d) −2 (e) 0 (f) 1

30. (a) ≈ 2.5 (b) 1 (c) −1 (d) Does not exist (e) 0 (f) 0

31.

-4 4

-4

4

–!

!

2

!

2

32.

-4 -2 2 4

-4

-2

2

4

–!

!

2

!

2

33.

-4 -2 2 4

-4

-2

2

4

–!

!

2

!

2

34.

-4 -2 2 4

-4

-2

2

4

–!

!

2

!

2

35.

-1 -0.5 0.5 1

-1

-0.5

0.5

1

–!

!

2

!

2

The limit does not exist.

36.

-0.5 0.5

-0.5

0.5

–!

!

2

!

2

The limit is 0.


37.

-0.5 0.5

-0.5

0.5

–!

!

2

!

2

The limit is −0.25.

38.-0.5 0.5

-3

-2

-1

–!

!

2

!

2

The limit is −3.

39.-0.5 0.5

-3

-2

-1

–!

!

2

!

2

The limit is −2.

40.

-0.5 0.5

-5

5

–!

!

2

!

2

The limit does not exist.

41. x → 1− 0.9 0.99 0.999 0.9999f(x) −3.25536642 −3.02276607 −3.00225263 −3.00022503

x → 1+ 1.1 1.01 1.001 1.0001f(x) −2.79817601 −2.97775903 −2.99775260 −2.99977503

limx→1

f(x) = −3

42. x → 1− 0.9 0.99 0.999 0.9999f(x) 1.05360516 1.00503359 1.00050033 1.00005000

x → 1+ 1.1 1.01 1.001 1.0001f(x) 0.95310180 0.99503309 0.99950033 0.99995000

limx→1

f(x) = 1

43. x → 0− −0.1 −0.01 −0.001 −0.0001f(x) −0.04995835 −0.00499996 −0.00050000 −0.00005000

x → 0+ 0.1 0.01 0.001 0.0001f(x) 0.04995835 0.00499996 0.00050000 0.00005000

limx→0

f(x) = 0

44. Since1− cosx

x2is an even function, it suffices to consider only x → 0+.

x → 0+ 0.1 0.01 0.001 0.0001f(x) 0.49958347 0.49999583 0.49999996 0.50000000

limx→0

f(x) = 0.5

45. Sincex

sin 3xis an even function, it suffices to consider only x → 0+.

x → 0+ 0.1 0.01 0.001 0.0001f(x) 0.33838634 0.33338334 0.33333383 0.33333334

2.2. LIMIT THEOREMS 77

limx→0

f(x) = 0.33333333

46. Sincetanx

xis an even function, it suffices to consider only x → 0+.

x → 0+ 0.1 0.01 0.001 0.0001f(x) 1.00334672 1.00003333 1.00000033 1.00000000

limx→0

f(x) = 1

47. x → 4− 3.9 3.99 3.999 3.9999f(x) 0.25158234 0.25015645 0.25001563 0.25000156

x → 4+ 4.1 4.01 4.001 4.0001f(x) 0.24845673 0.24984395 0.24998438 0.24999844

limx→4

f(x) = 0.25

48. x → 3− 2.9 2.99 2.999 2.9999f(x) −0.52186477 −0.50209311 −0.50020843 −0.50002083

x → 3+ 3.1 3.01 3.001 3.0001f(x) −0.48008703 −0.49792633 −0.49979176 −0.49997917

limx→3

f(x) = −0.5

49. x → 1− 0.9 0.99 0.999 0.9999f(x) 4.43900000 4.94039900 4.99400400 4.99940004

x → 1+ 1.1 1.01 1.001 1.0001f(x) 5.64100000 5.06040010 5.00600400 5.00060004

limx→1

f(x) = 5

50. x → −2− −2.1 −2.01 −2.001 −2.0001f(x) 12.61000000 12.06010000 12.00600100 12.00060001

x → −2+ −1.9 −1.99 −1.999 −1.9999f(x) 11.41000000 11.94010000 11.99400100 11.99940001

limx→−2

f(x) = 12

2.2 Limit Theorems

1. 15

2. cosπ = −1

3. −12

4. −3

5. 4


6. −125

7. 4

8. −136

9. −8/5

10. does not exist

11. 14

12. 4

13. 28/9

14. limx→6

x2 − 6x

x2 − 7x+ 6= lim

x→6

x(x− 6)

(x− 1)(x− 6)= lim

x→6

x

x− 1=

6

5

15. −1

16. 16

17.√7

18. 3

19. does not exist

20. 16

21. limy→−5

y2 − 25

y + 5= lim

y→−5(y − 5) = −10

22. limu→8

u2 − 5u− 24

u− 8= lim

u→8(u+ 3) = 11

23. limx→1

x3 − 1

x− 1= lim

x→1

(x− 1)(x2 + x+ 1)

x− 1= lim

x→1(x2 + x+ 1) = 3

24. limt→−1

t3 + 1

t2 − 1= lim

t→−1

(t+ 1)(t2 − t+ 1)

(t+ 1)(t− 1)= lim

t→−1

t2 − t+ 1

t− 1= −3

2

25. limx→10

(x− 2)(x+ 5)

x− 8=

8(15)

2= 60

26. limx→−3

2x+ 6

4x2 − 36= lim

x→−3

2(x+ 3)

4(x+ 3)(x− 3)= lim

x→−3

1

2(x− 3)= − 1

12

27. limx→2

x3 + 3x2 − 10x

x− 2= lim

x→2

x(x+ 5)(x− 2)

x− 2= lim

x→2x(x+ 5) = 14

28. limx→1.5

2x2 + 3x− 9

x− 1.5= lim

x→1.5

(2x− 3)(x+ 3)

x− 1.5= lim

x→1.5

2(x− 1.5)(x+ 3)

x− 1.5= lim

x→1.52(x+ 3) = 9

2.2. LIMIT THEOREMS 79

29. limt→1

t3 − 2t+ 1

t3 + t2 − 2= lim

t→1

(t− 1)(t2 + t− 1)

(t− 1)(t2 + 2t+ 2)= lim

t→1

t2 + t− 1

t2 + 2t+ 2=

1

5

30. limx→0

x3(x4 + 2x3)−1 = limx→0

x3

x4 + 2x3= lim

x→0

1

x+ 2=

1

2

31. limx→0+

(x+ 2)(x5 − 1)3

(√x+ 4)2

=2(−1)

16= −1

8

32. limx→−2

x√x+ 4 3

√x− 6 = −2

√2 3√−8 = 4

√2

33. limx→0

�x2 + 3x− 1

x+

1

x

�= lim

x→0

x2 + 3x

x= lim

x→0(x+ 3) = 3

34. limx→2

�1

x− 2− 6

x2 + 2x− 8

�= lim

x→2

�1

x− 2− 6

(x− 2)(x+ 4)

�

= limx→2

�x+ 4

(x− 2)(x+ 4)− 6

(x− 2)(x+ 4)

�

= limx→2

x− 2

(x− 2)(x+ 4)= lim

x→2

1

x+ 4=

1

6

35. does not exist

36. −210 or −1024

37. 2

38.2√2

3√4

39. limh→4

�h

h+ 5

�h2 − 16

h− 4

�2= lim

h→4

�h

h+ 5(h2 + 8h+ 16) =

128

3

40. 16

41. limx→0−

5

�x3 − 64x

x2 + 2x= lim

x→0−

5

�x2 − 64

x+ 2= −2

42. −100, 000

43. a2 − 2ab+ b2

44. limx→−1

√u2x2 + 2xu+ 1 = lim

x→−1

√u2 − 2u+ 1 = lim

x→−1

�(u− 1)2 = |u− 1|

45. limh→0

(8 + h)2 − 64

h= lim

h→0

16h+ h2

h= lim

h→0(16 + h) = 16

46. limh→0

1

h[(1 + h)3 − 1] = lim

h→0(h2 + 3h+ 3) = 3


47. limh→0

1

h

�1

x+ h− 1

x

�= lim

h→0

1

h

�x− (x+ h)

(x+ h)x

�= lim

h→0

−h

hx(x+ h)

= limh→0

− 1

x2 + hx= − 1

x2

48. limh→0

√x+ h−√

x

h= lim

h→0

√x+ h−√

x

h

√x+ h+

√x√

x+ h+√x

= limh→0

(x+ h)− x

h(√x+ h+

√x)

= limh→0

1√x+ h+

√x=

1

2√x

49. limt→1

√t− 1

t− 1= lim

t→1

√t− 1

t− 1

√t+ 1√t+ 1

= limt→1

1√t+ 1

=1

2

50. limu→5

√u+ 4− 3

u− 5= lim

u→5

√u+ 4− 3

u− 5

√u+ 4 + 3√u+ 4 + 3

= limu→5

u− 5

(u− 5)(√u+ 4 + 3)

= limu→5

1√u+ 4 + 3

=1

6

51. limv→0

√25 + v − 5√1 + v − 1

= limv→0

�√25 + v − 5√1 + v − 1

√25 + v + 5√1 + v + 1

� √1 + v + 1√25 + v + 5

= limv→0

v

v

√1 + v + 1√25 + v + 5

=1

5

52. limx→1

4−√x+ 15

x2 − 1= lim

x→1

4−√x+ 15

x2 − 1

4 +√x+ 15

4 +√x+ 15

= limx→1

1− x

(x+ 1)(x− 1)(4 +√x+ 15)

= limx→1

−(x− 1)

(x+ 1)(x− 1)(4 +√x+ 15)

= limx→1

−1

(x+ 1)(4 +√x+ 15)

= − 1

16

53. 32

54. 64

55.1

2

56.

�4

2=

√2

57. does not exist

58. 8

59. 8a

60.3

2

2.3. CONTINUITY 81

61. (a) limx→1

x100 − 1

x2 − 1= lim

x→1

x100 − 1

(x+ 1)(x− 1)

= limx→1

1

x+ 1· x

100 − 1

x− 1=

1

2· 100 = 50

(b) limx→1

x50 − 1

x− 1= lim

x→1

x50 − 1

x− 1· x

50 + 1

x50 + 1

= limx→1

x100 − 1

x− 1· 1

x50 + 1= 100 · 1

2= 50

(c) limx→1

(x100 − 1)2

(x− 1)2= lim

x→1

x100 − 1

x− 1· x

100 − 1

x− 1= 100 · 100 = 10, 000

62. (a) limx→0

2x

sinx= 2 lim

x→0

1�sinx

x

� = 2 ·limx→0

1

limx→0

sinx

x

= 2

(b) limx→0

1− cos2 x

x2= lim

x→0

sin2 x

x2= lim

x→0

sinx

x· sinx

x= 1 · 1 = 1

(c) limx→0

8x2 − sinx

x= lim

x→0

�8x− sinx

x

�= lim

x→08x− lim

x→0

sinx

x= −1

63. limx→0

sinx = limx→0

�x · sinx

x

�= lim

x→0x · lim

x→0

sinx

x= 0 · 1 = 0

64. limx→2

[2f(x)− 5] = limx→2

(x+ 3)

�2f(x)− 5

x+ 3

�= 5 · 4 = 20

2 limx→2

f(x)− limx→2

5 = 20

limx→2

f(x) =20 + 5

2= 12.5

2.3 Continuity

1. Continuous everywhere


3. Discontinuous at 3 and 6

4. Discontinuous at −1 and 1

5. Discontinuous atnπ

2, for n = 0, 1, 2, . . .

6. Discontinuous at −3 andπ

2+ nπ, for n an integer

7. Discontinuous at 2




10. Discontinuous at x < 0 and1

2

11. Discontinuous at e−2


13. (a) yes (b) yes

14. (a) no (b) yes

15. (a) yes (b) yes

16. (a) yes (b) yes

17. (a) no (b) no

18. (a) yes (b) yes

19. (a) yes (b) no

20. (a) no (b) no

21. Solving 2 + secx = 0, we obtain cosx = − 12 , so x = 2π

3 + 2nπ or x = 4π3 + 2nπ. Thus, f(x)

is discontinuous on (−∞,∞) and on [π2 ,3π2 ].

22. Since sin 1x is discontinuous only at x = 0, it is continuous on [ 1π ,∞) and discontinuous on

[−2π , 2

π ].

23. Since f(x) is discontinuous only at x = 2, it is discontinuous on [−1, 3] and continuous on(2, 4].

24. Since f(x) is defined and continuous exactly on (1, 5], it is continuous on [2, 4] and discontin-uous on [1, 5].

25. Since limx→4−

f(x) = 4m and limx→4+

f(x) = 16, we have 4m = 16 and m = 4.

26. Since limx→2

f(x) = limx→2

(x− 2)(x+ 2)

x− 2= 4 we have f(2) = m and m = 4.


f(x) = 3m, limx→3+

f(x) = 3, and f(3) = n, we have 3m = 3 = n, so m = 1 and

n = 3.


f(x) = m − n, limx→1+

f(x) = 2m + n, and f(1) = 5, we have m − n = 5 and

2m+ n = 5. Adding, we obtain 3m = 10, so m = 10/3 and n = −5/3.

2.3. CONTINUITY 83

29. Discontinuous atn

2, n an integer

-3 3

-3

3

30. Discontinuous at every integer

-3 3

-3

3

31. Since limx→9

x− 9√x− 3

= limx→9

(√x+ 3)(

√x− 3)√

x− 3= lim

x→9(√x+ 3) = 6, define f(9) = 6.

32. Since limx→1

x4 − 1

x2 − 1= lim

x→1

(x2 + 1)(x2 − 1)

x2 − 1= lim

x→1(x2 + 1) = 2, define f(1) = 2.

33. limx→π/6

sin(2x+π

3) = sin

�lim

x→π/6(2x+

π

3)

�= sin

2π

3=

√3

2

34. limx→π2

cos√x = cos

�lim

x→π2

√x

�= cosπ = −1

35. limx→π/2

sin(cosx) = sin

�lim

x→π/2cosx

�= sin(cos

π

2) = sin 0 = 0

36. limx→π/2

[1 + cos(cosx)] = 1 + cos

�lim

x→π/2cosx

�= 1 + cos(cos

π

2) = 1 + cos 0 = 2

37. limt→π

cos

�t2 − π2

t− π

�= cos

�limt→π

(t− π)(t+ π)

t− π

�= cos 2π = 1

38. limt→0

tan

�πt

t2 + 3t

�= tan

�limt→0

πt

t(t+ 3)

�= tan

�limt→0

π

t+ 3

�= tan

π

3=

√3

39. limt→π

√t− π + cos2 t =

√cos2 π =

√1 = 1

40. limt→1

(4t+ sin 2πt)3 =�limt→1

(4t+ sin 2πt)�3

= (4 + sin 2π)3 = 43 = 64


41. limx→−3

sin−1

�x+ 3

x2 + 4x+ 3

�= sin−1

�lim

x→−3

x+ 3

x2 + 4x+ 3

�

= sin−1

�lim

x→−3

x+ 3

(x+ 3)(x+ 1)

�

= sin−1

�lim

x→−3

1

x+ 1

�= sin−1

�−1

2

�= −π

6

42. limx→π

ecos 3x = elimx→π

cos 3x= ecos 3π = e−1

43. Since (f ◦ g)(x) = 1√x+ 3

, f ◦ g is continuous for x+ 3 > 0 or on (−3,∞).

44. Since (f◦g)(x) = 5(x− 2)2

(x− 2)2 − 1=

5(x− 2)2

x2 − 4x+ 3=

5(x− 2)2

(x− 1)(x− 3), we see that f◦g is continuous

for x �= 1 and x �= 3 or on (∞, 1) ∪ (1, 3) ∪ (3,∞).

45. f(1) = −1, f(5) = 15. By the Intermediate Value Theorem, since −1 ≤ 8 ≤ 15, there existsc ∈ [1, 5] such that c2−2c = 8. Setting c2−2c−8 = 0 gives us (c−4)(c+2) = 0 or c = −2, 4.On [1, 5], c = 4.

46. f(−2) = 3, f(3) = 13. By the Intermediate Value Theorem, since 3 ≤ 6 ≤ 13, there exists

c ∈ [−2, 3] such that c2+c+1 = 6. Setting c2+c−5 = 0 gives us c =−1±

�1− 4(1)(−5)

2=

−1±√21

2. On [−2, 3], c =

−1 +√21

2.

47. f(−2) = −3, f(2) = 5. By the Intermediate Value Theorem, since −3 ≤ 1 ≤ 5, there existsc ∈ [−2, 2] such that c3 − 2x+ 1 = 1. Setting c3 − 2c = 0 gives us c(c2 − 2) = 0. On [−2, 2],c = 0,±

√2.

48. f(0) = 10, f(1) = 5. By the Intermediate Value Theorem, since 5 ≤ 8 ≤ 10, there exists

c ∈ [0, 1] such that10

c2 + 1= 8. Setting c2 + 1 =

5

4or c2 − 1

4= 0 gives us (c+

1

2)(c− 1

2) = 0

or c = ±1

2. On [0, 1], c =

1

2.

49. Since f(0) = −7, f(3) = 242, and −7 ≤ 50 ≤ 242, then by the Intermediate Value Theoremthere exists c ∈ [0, 3] such that f(c) = 50.

50. Since f(a) > g(a), then (f − g)(a) > 0. Since f(b) < g(b), then (f − g)(b) < 0. By thecorollary to the Intermediate Value Theorem, there exists c ∈ (a, b) such that (f − g)(c) = 0.Then f(c) = g(c).

51. The equation will have a solution on (0, 1) if f(x) = 2x7+x−1 is 0 on (0, 1). Since f(0) = −1and f(1) = 2, then by the Intermediate Value Theorem f(c) = 0 for some c ∈ (0, 1).

52. Let f(x) =x2 + 1

x+ 3+

x4 + 1

x− 4. Then f(0) =

1

3− 1

4> 0 and f(1) =

1

2− 2

3< 0. Thus, by the

corollary to the Intermediate Value Theorem, f(c) = 0 for some c between 0 and 1, and hencebetween −3 and 4.

2.3. CONTINUITY 85

53. Let f(x) = e−x − lnx. Then f(1) = e−1 − ln 1 = e−1 > 0 and f(2) = e−2 − ln2 < 0. Thus,by the corollary to the Intermediate Value Theorem, f(c) = 0 for some c ∈ (1, 2).

54. Since

�sin

π

2

�

�π2

� =2

π,sinπ

π= 0, and 0 ≤ 1

2≤ 2

π, then by the Intermediate Value Theorem,

there exists c betweenπ

2and π such that

sinx

x=

1

2.

-3 3

-3

3

55. In [−2,−1] the zero is approximately −1.21. In [−1, 0] the zero is approx-imately −0.64. In [1, 2] the zero is approximately 1.34.

-3 3

-3

3

56. In [0, 1] the zero is approximately 0.75.

57. We want to solve f(x) = x5 + 2x − 7 = 50 or x5 + 2x − 57 = 0. It is easily seen that theexpression on the left side of this equation is negative when x = 2 and positive when x = 3.Applying the bisection method on [2, 3], we find c ≈ 2.21.

58. Applying the bisection method to f(x) = 2x7 + x− 1 on [0, 1], we find c ≈ 0.75.

59. In the solution of Problem 52 we saw that there is a zero in [0, 1]. Applying the bisectionmethod on this interval, we find c ≈ 0.78.

60. (a) If h is the height of the cylinder, then the volume is given by V = πr2h and the surface

area is S = 2πr2 + 2πrh. Solving the latter equation for h, we obtain h =S

2πr− r.

Substituting into the formula for V , we find V =1

2Sr − πr3 or 2πr3 − Sr + 2V = 0.

(b)

10 20

-5000

5000

(c) From the graph, we observe zeros in [3, 4] and [14, 15]. The bisection method gives

r ≈ 3.48 ft and r ≈ 14.91 ft. The corresponding values of h are h =1800

2πr− r ≈ 78.84 ft

and h =1800

2πr− r ≈ 4.29 ft.


61. Since f and g are continuous at a, then limx→a

f(x) = f(a) and limx→a

g(x) = g(a). From this, we

get:

limx→a

(f + g)(x) = limx→a

[f(x) + g(x)] = limx→a

f(x) + limx→a

g(x)

= f(a) + g(a) = (f + g)(a)

Thus, f + g is continuous at a.

62. Since f and g are continuous at a, then limx→a

f(x) = f(a) and limx→a

g(x) = g(a). From this we

get:

limx→a

(f/g)(x) = limx→a

[f(x)/g(x)] = limx→a

f(x)/ limx→a

g(x)

= f(a)/g(a) = (f/g)(a) (since g(a) �= 0)

Thus, f/g is continuous at a.

63. f ◦ g will be discontinuous whenever cosx is an integer. In the interval [0, 2π), this will bethe case whenever x = 0, π/2, π, or 3π/2. Thus, f ◦ g will be discontinuous for x = nπ/2, nan integer.

-3 3

-3

364. (f ◦ g)(x) =�

|x+ 1|, x < 0|x− 1|, x ≥ 0

is continuous at x = 0.

-3 3

-3

3

(g ◦ f)(x) = |x|− 1 is continuous at x = 0.

65. (a) For any real a, limx→a

f(x) does not exist since f takes on the values 0 and 1 arbitrarily

close to any real number. Therefore, the Dirichlet function is discontinuous at every realnumber.

(b) The graph consists of infinitely many points on each of the lines y = 0 and y = 1. Infact, between any two real numbers, there are infinitely many points of the graph on theline y = 1 and infinitely many points of the graph on the line y = 0.

(c) Let r be a positive rational number. If x is rational, then x + r is rational so thatf(x+r) = 1 = f(x). If x is irrational, then x+r is irrational so that f(x+r) = 0 = f(x).

2.4 Trigonometric Limits

1. limt→0

sin 3t

2t=

1

2limt→0

sin 3t

t=

3

2

2.4. TRIGONOMETRIC LIMITS 87

2. limt→0

sin(−4t)

t= −4

3. limx→0

sinx

4 + cosx=

0

4 + 1= 0

4. limx→0

1 + sinx

1 + cosx=

1 + 0

1 + 1=

1

2

5. limx→0

cos 2x

cos 3x= 1

6. limx→0

tanx

3x=

1

3limx→0

�sinx

x· 1

cosx

�=

1

3(1 · 1) = 1

3

7. limt→0

1

t sec t csc 4t= lim

t→0

�sin 4t

t· cos t

�= 4 · 1 = 4

8. limt→0

(5t cot 2t) = 5 limt→0

t cos 2t

sin 2t= 5 lim

t→0

�cos 2t · 1

(sin 2t)/t

�

= 5�limt→0

cos 2t��

limt→0

1

(sin 2t)/t

�= 5 · 1 · 1

2=

5

2

9. limt→0

2 sin2 t

t cos2 t= 2 lim

t→0

�sin t

t· sin t

cos2 t

�= 2 · 1 · 0 = 0

10. limt→0

sin2(t/2)

sin t= lim

t→0

�sin(t/2)

t· t sin(t/2)

sin t

�

=

�limt→0

sin(t/2)

t

� �limt→0

sin(t/2)

(sin t)/t

�=

1

2· 01= 0

11. limt→0

sin2 6t

t2= lim

t→0

�sin 6t

t

�2

= 62 = 36

12. limt→0

t3

sin2 3t= lim

t→0

�t · t2

sin2 3t

�=

�limt→0

t��

limt→0

1

[(sin 3t)/t]2

�= 0 · 1

32= 0

13. limx→1

sin(x− 1)

2x− 2=

1

2limx→1

sin(x− 1)

x− 1=

1

2

14. limx→2π

x− 2π

sinx= lim

x→2π

x− 2π

sin(x− 2π)= lim

x→2π

1

sin(x− 2π)/(x− 2π)= 1

15. limx→0

cosx

xdoes not exist.

16. limθ→π/2

1 + sin θ

cos θdoes not exist.

17. limx→0

cos(3x− π/2)

x= lim

x→0

sin 3x

x= 3


18. limx→−2

sin(5x+ 10)

4x+ 8=

5

5· 14· limx→−2

sin(5x+ 10)

x+ 2=

5

4lim

x→−2

sin(5x+ 10)

5x+ 10=

5

4

19. limt→0

sin 3t

sin 7t= lim

t→0

�sin 3t

t· t

sin 7t

�=

�limt→0

sin 3t

t

��limt→0

1

(sin 7t)/t

�= 3 · 1

7=

3

7

20. limt→0

sin 2t csc 3t = limt→0

sin 2t

sin 3t= lim

t→0

�sin 2t

t· t

sin 3t

�

=

�limt→0

sin 2t

t

��limt→0

1

(sin 3t)/t

�= 2 · 1

3=

2

3

21. limt→0+

sin t√t

= limt→0+

�√t · sin t

t

�=

�limt→0+

√t

��limt→0+

sin t

t

�= 0 · 1 = 0

22. Letting u =√t, we have lim

t→0

1− cos√t√

t= lim

u→0

1− cosu

u= 0.

23. limt→0

t2 − 5t sin t

t2= lim

t→0

�1− 5

�sin t

t

��= 1− 5 = −4

24. limt→0

cos 4t

cos 8t=

1

1= 1

25. limx→0+

(x+ 2√sinx)2

x= lim

x→0+

x2 + 4x√sinx+ 4 sinx

x

= limx→0+

�x+ 4

√sinx+

4 sinx

x

�= 0 + 0 + 4 = 4

26. limx→0

(1− cosx)2

x=

�limx→0

(1− cosx)��

limx→0

1− cosx

x

�= 0 · 0 = 0

27. limx→0

cosx− 1

cos2 x− 1=

�limx→0

cosx− 1

cosx− 1

��limx→0

1

cosx+ 1

�= 1 · 1

2=

1

2

28. limx→0

sinx+ tanx

x= lim

x→0

�sinx

x+

�sinx

x· 1

cosx

��= 1 + (1 · 1) = 2

29. Letting u = x2, we have limx→0

sin 5x2

x2= lim

u→0

sin 5u

u= 5.

30. limt→0

t2

1− cos t= lim

t→0

�t2

1− cos t· 1 + cos t

1 + cos t

�= lim

t→0

�t2

sin2 t· (1 + cos t)

�

=

�limt→0

1

(sin t)/t

�2 �limt→0

(1 + cos t)�= 12 · 2 = 2

31. First, rewrite limx→2

sin(x− 2)

x2 + 2x− 8as lim

x→2

sin(x− 2)

(x− 2)(x+ 4).

Letting u = x− 2, we get limu→0

�sinu

u· 1

u+ 6

�= 1 · 1

6=

1

6.


32. First, rewrite limx→3

x2 − 9

sin(x− 3)as lim

x→3

(x− 3)(x+ 3)

sin(x− 3). Letting u = x− 3:

limu→0

� u

sinu· (u+ 6)

�= lim

u→0

�1

(sinu)/u· (u+ 6)

�=

1

1· 6 = 6

33. limx→0

2 sin 4x+ 1− cosx

x= lim

x→0

�2 sin 4x

x+

1− cosx

x

�= 8 + 0 = 8

34. limx→0

4x2 − 2 sinx

x= lim

x→0

�4x− 2 sinx

x

�= 0− 2 = −2

35. Start by multiplying the function by1 + tanx

1 + tanx, producing:

limx→π/4

1− tanx

cosx− sinx= lim

x→π/4

�1− tanx

cosx− sinx· 1 + tanx

1 + tanx

�

= limx→π/4

1− tan2 x

(cosx− sinx)(1 + tanx)

Focusing first on the denominator, we multiply out and simplify:

(cosx− sinx)(1 + tanx) = cosx+ cosx tanx− sinx− sinx tanx

= cosx+ cosx

�sinx

cosx

�− sinx

�cosxcosx

�− sinx

�sinx

cosx

�

= cosx− sin2 x

cosx=

cos2 x− sin2 x

cosxSubstituting this result back into the function, we get:

1− tan2 x

(cosx− sinx)(1 + tanx)= (1− tan2 x)

�cosx

cos2 x− sin2 x

�

=

cosx− cosx

�sin2 x

cos2 x

�

cos2 x− sin2 x=

cosx−�sin2 x

cosx

�

cos2 x− sin2 x

=

�cos2 x− sin2 x

cosx

��1

cos2 x− sin2 x

�=

1

cosx

Finally, returning to the limit, we have:

limx→π/4

1− tanx

cosx− sinx= lim

x→π/4

1

cosx=

1√2/2

=√2

36. Using the trigonometric identity cos 2x = cos2 x− sin2 x, we have:

limx→π/4

cos 2x

cosx− sinx= lim

x→π/4

cos2 x− sin2 x

cosx− sinx

= limx→π/4

(cosx+ sinx)(cosx− sinx)

cosx− sinx

= limx→π/4

(cosx+ sinx) =√2


37. limh→0

f�π4+ h

�− f

�π4

�

h= lim

h→0

sin�π4+ h

�− sin

�π4

�

h

= limh→0

sin(π/4) cosh+ cos(π/4) sinh− sin(π/4)

h

= limh→0

(√2/2) cosh+ (

√2/2) sinh− (

√2/2)

h

=

√2

2limh→0

cosh+ sinh− 1

h=

√2

2limh→0

�cosh− 1

h+

sinh

h

�=

√2

2

38. limh→0

f�π6+ h

�− f

�π6

�

h= lim

h→0

cos�π6+ h

�− cos

�π6

�

h

= limh→0

cos(π/6) cosh− sin(π/6) sinh− cos(π/6)

h

= limh→0

(√3/2) cosh− (1/2) sinh− (

√3/2)

h

= limh→0

�√3

2

�cosh− 1

h

�− 1

2

�sinh

h

��=

√3

2· 0− 1

2· 1 = −1

2

39. Since −1 ≤ sin1

x≤ 1, then −|x| ≤ x sin

1

x≤ |x|. Since lim

x→0(−|x|) = 0 and lim

x→0|x| = 0, then

by the Squeeze Theorem, limx→0

x sin1

x= 0.

40. Since −1 ≤ cosπ

x≤ 1, then −x2 ≤ x2 cos

π

x≤ x2. Since lim

x→0−x2 = 0 and lim

x→0x2 = 0, then

by the Squeeze Theorem, limx→0

x2 cosπ

x= 0.

41. For both limits, we use the result from Problem 39, limx→0

x sin1

x= 0:

(a) limx→0

x3 sin1

x= lim

x→0

�x2 · x sin 1

x

�= lim

x→0x2 · lim

x→0x sin

1

x= 0 · 0 = 0

(b) limx→0

x2 sin21

x=

�limx→0

x sin1

x

��limx→0

x sin1

x

�= 0 · 0 = 0

42. |f(x)| ≤ B means that B ≥ 0 and therefore −B ≤ f(x) ≤ B. Thus, −Bx2 ≤ x2f(x) ≤ Bx2

in that interval. Since limx→0

(−Bx2) = 0 and limx→0

Bx2 = 0, then by the Squeeze Theorem,

limx→0

x2f(x) = 0.

43. Since limx→2

(2x−1) = 3 and limx→2

(x2−2x+3) = 3, then by the Squeeze Theorem, limx→2

f(x) = 3.

44. Since |f(x) − 1| ≤ x2, then f(x) − 1 ≤ x2, or f(x) ≤ x2 + 1 when f(x) − 1 > 0. However,f(x) ≤ x2+1 is in fact true for all x, since x2 ≥ 0 for all x. Similarly, we have −x2 ≤ f(x)−1,or −x2+1 ≤ f(x) for all x. Since lim

x→0(−x2+1) = 1 and lim

x→0(x2+1) = 1, then by the Squeeze

Theorem, limx→0

f(x) = 1.


45. Let t = x− π

4. Thus, x = t+

π

4and we have the following substitutions:

sinx = sin(t+π

4) = sin t cos

π

4+ cos t sin

π

4=

√2

2sin t+

√2

2cos t

cosx = cos(t+π

4) = cos t cos

π

4− sin t sin

π

4=

√2

2cos t−

√2

2sin t

sinx− cosx =

�√2

2sin t+

√2

2cos t

�−�√

2

2cos t−

√2

2sin t

�=

√2 sin t

With these substitutions, limx→π/4

sinx− cosx

x− (π/4)= lim

t→0

√2 sin t

t=

√2.

46. Let t = x− π. Thus, x = t+ π. Substituting, we get:

limx→π

x− π

tan 2x= lim

t→0

t

tan(2t+ 2π)= lim

t→0

t

tan 2t

= limt→0

1�tan 2t

t

� = limt→0

11

cos 2t· sin 2t

t

=1

1 · 2 =1

2

47. Let t = π − (π/x). Therefore π/x = π − t and sin(π/x) = sin(π − t) = sin t. In addition, we

can derive x− 1 =t

π − t, giving us:

limx→1

sin(π/x)

x− 1= lim

t→0

(sin t)(π − t)

t= lim

t→0

sin t

t· limt→0

(π − t) = 1 · π = π

48. Let t =π

2− π

x. Substituting in the same way as in Problem 47, we get:

limx→2

cos(π/x)

x− 2= lim

t→0

(sin t)(π − 2t)

4t= lim

t→0

sin t

t· limt→0

π − 2t

4=

π

4

49. f is continuous at x = 0 because limx→0

sinx

x= 1 = f(0).

50. Since |x| =�

x, x > 0−x, x < 0

, knowing that limx→0

sinx

x= 1 means:

limx→0+

sin |x|x

= limx→0+

sinx

x= 1

limx→0−

sin |x|x

= limx→0−

sin(−x)

x= lim

x→0−

− sinx

x= − lim

x→0−

sinx

x= −1

Since limx→0+

sin |x|x

�= limx→0−

sin |x|x

, then limx→0

sin |x|x

does not exist.


2.5 Limits that Involve Infinity

1. −∞ 2. ∞ 3. ∞ 4. −∞5. ∞ 6. −∞ 7. ∞ 8. −∞

9. limx→∞

x2 − 3x

4x2 + 5= lim

x→∞

1− 3/x

4 + 5/x2=

1

4

10. limx→∞

x2

1 + x−2= lim

x→∞

1

1/x2 + 1/x4= ∞

11. 5

12. limx→−∞

�63√x+

15√x

�= 0

13. limx→∞

8−√x

1 + 4√x= lim

x→∞

(8/√x)− 1

(1/√x) + 4

= −1

4

14. limx→−∞

1 + 7 3√x

2 3√x

= limx→−∞

1/ 3√x+ 7

2=

7

2

15. limx→∞

�3x

x+ 2− x− 1

2x+ 6

�= lim

x→∞

�3

1 + 2/x− 1− 1/x

2 + 6/x

�= 3− 1

2=

5

2

16. limx→∞

�x

3x+ 1

��4x2 + 1

2x2 + x

�3

= limx→∞

�1

3 + 1/x

��4 + 1/x2

2 + 1/x

�3

=1

3· 23 =

8

3

17. limx→∞

�3x+ 2

6x− 8= lim

x→∞

�3 + 2/x

6− 8/x=

�1

2=

√2

2

18. limx→−∞

3

�2x− 1

7− 16x= lim

x→−∞3

�2− 1/x

7/x− 16= 3

�− 2

16= −1

2

19. limx→∞

�x−

√x2 + 1

�= lim

x→∞

�x−

√x2 + 1

�· x+

√x2 + 1

x+√x2 + 1

= limx→∞

−1

x+√x2 + 1

= 0

20. limx→∞

��x2 + 5x− x

�= lim

x→∞

��x2 + 5x− x

�·√x2 + 5x+ x√x2 + 5x+ x

= limx→∞

5x√x2 + 5x+ x

= limx→∞

5�1 + 5/x+ 1

=5

2

21. limx→∞

cos

�5

x

�= cos

�limx→∞

�5

x

��= 1

22. limx→−∞

sin

�πx

3− 6x

�= lim

x→−∞sin

�π

3/x− 6

�= sin

�lim

x→−∞

�π

3/x− 6

��= −1

2

2.5. LIMITS THAT INVOLVE INFINITY 93

23. limx→−∞

sin−1

�x√

4x2 + 1

�= lim

x→−∞sin−1

�x

|x|

�

�4 + 1/x2

= limx→−∞

sin−1

�x

−x

�

�4 + 1/x2

= sin−1

�lim

x→−∞

�−1�

4 + 1/x2

��= sin−1

�−1

2

�= −π

6

24. limx→∞

ln

�x

x+ 8

�= lim

x→∞ln

�1

1 + 8/x

�= ln

�limx→∞

�1

1 + 8/x

��= ln 1 = 0

25. Start with4x+ 1√x2 + 1

=

�4x

|x| +1

|x|

�

�1 + 1/x2

. From this, limx→−∞

f(x) = limx→−∞

−4− 1/x�1 + 1/x2

= −4 and

limx→∞

f(x) = limx→∞

4 + 1/x�1 + 1/x2

= 4.

26. Start with

√9x2 + 6

5x− 1=

�9 + 6/x2

�5x

|x| −1

|x|

� . From this, limx→−∞


�9 + 6/x2

−5 + 1/x=

√9

−5=

−3

5and lim

x→∞f(x) = lim

x→∞

�9 + 6/x2

5− 1/x=

√9

5=

3

5.

27. Start with2x+ 1√3x2 + 1

=

�2x

|x| +1

|x|

�

�3 + 1/x2



−2− 1/x�3 + 1/x2

= − 2√3=

−2√3

3and lim

x→∞f(x) = lim

x→∞

2 + 1/x�3 + 1/x2

=2√3=

2√3

3.

28. Start with−5x2 + 6x+ 3√

x4 + x2 + 1=

�−5 +

6

|x| +3

x2

�

�1 + 1/x2 + 1/x4



−5− 6/x+ 3/x2

�1 + 1/x2 + 1/x4

=

−5√1= −5 and lim

x→∞f(x) = lim

x→∞

−5 + 6/x+ 3/x2

�1 + 1/x2 + 1/x4

=−5√1= −5.

29. limx→−∞

ex − e−x

ex + e−x=

�lim

x→−∞ex�−

�lim

x→−∞e−x

�

�lim

x→−∞ex�+

�lim

x→−∞e−x

� =

0−�

limx→−∞

e−x

�

0 +

�lim

x→−∞e−x

�

= limx→−∞

−e−x

e−x= lim

x→−∞−1 = −1

limx→∞

ex − e−x

ex + e−x=

�limx→∞

ex�−�limx→∞

e−x�

�limx→∞

ex�+�limx→∞

e−x� =

�limx→∞

ex�− 0

�limx→∞

ex�+ 0

= limx→∞

ex

ex= lim

x→∞1 = 1


30. limx→−∞

�1 +

2e−x

ex + e−x

�= 1 +

limx→−∞

2e−x

�lim

x→−∞ex�+

�lim

x→−∞e−x

�

= 1 +lim

x→−∞2e−x

0 +

�lim

x→−∞e−x

� = 1 + limx→−∞

2e−x

e−x= 1 + lim

x→−∞2 = 3

limx→∞

�1 +

2e−x

ex + e−x

�= 1 +

limx→∞

2e−x

�limx→∞

ex�+�limx→∞

e−x� = 1 + lim

x→∞

0

ex= 1

31. limx→−∞

|x− 5|x− 5

= limx→−∞

−x+ 5

x− 5= lim

x→−∞

−1 + 5/x

1− 5/x= −1

limx→∞

|x− 5|x− 5

= limx→∞

x− 5

x− 5= 1

32. limx→−∞

|4x|+ |x− 1|x

= limx→−∞

−4x− (x− 1)

x= lim

x→−∞

−5x+ 1

x

= limx→−∞

−5 + 1/x

1= −5

limx→∞

|4x|+ |x− 1|x

= limx→∞

4x+ x− 1

x= lim

x→∞

5x− 1

x= lim

x→∞

5− 1/x

1= 5

33.-5 5

-5

5

Vertical asymptote: noneHorizontal asymptote: y = 0

34.-5 5

-5

5


35. -5 5

-5

5

Vertical asymptote: x = −1Horizontal asymptote: none


36.-5 5

-5

5

Vertical asymptote: x = −1Horizontal asymptote: y = 1

37.-5 5

-5

5

Vertical asymptote: x = 0, x = 2Horizontal asymptote: y = 0

38.-5 5

-5

5


39.-5 5

-5

5

Vertical asymptote: x = 1Horizontal asymptote: y = 1

40. 10

-10

10

Vertical asymptote: x = 0Horizontal asymptote: y = −1

41.-10 10

-10

10

Vertical asymptote: noneHorizontal asymptote: y = −1, y = 1

42.-10 10

-10

10

Vertical asymptote: x = −1, x = 1Horizontal asymptote: y = −1, y = 1


43. (a) 2 (b) −∞ (c) 0 (d) 2

44. (a) ∞ (b) ∞ (c) 1 (d) 3

45. (a) −∞ (b) −3/2 (c) ∞ (d) 0

46. (a) ∞ (b) −∞ (c) 0 (d) 0

47.-5 5

-5

5

48.-5 5

-5

5

49.-5 5

-5

5

50.-5 5

-5

5

51. limx→∞

x sin3

x= lim

x→∞

�x sin

3

x

��3/x

3/x

�= lim

x→∞x(3/x)

�sin 3/x

3/x

�

=

�limx→∞

x · 3x

��limx→∞

sin 3/x

3/x

�=

�limx→∞

3��

limx→∞

sin 3/x

3/x

�

At this point, we substitute t = 3/x, resulting in:

�limx→∞

3��

limx→∞

sin 3/x

3/x

�= 3 lim

t→0

sin t

t= 3

52. limv→c−

m0�1− v2/c2

= limv→c−

m0√1− 1

= limv→c−

m0

0; so as v → c−, m → ∞.


53. x → ∞ 10 100 1000 10000f(x) 1.99986667 1.99999999 2.00000000 2.00000000

limx→∞

x2 sin2

x2= 2

54. x → ∞ 10 100 1000 10000f(x) 0.95114995 0.99501240 0.99950012 0.99995000

limx→∞

�cos

1

x

�x

= 1

55.-5 5

-5

5

(a) limx→−1+

f(x) = ∞ (b) limx→0

f(x) ≈ 2.7 (c) limx→∞

f(x) = 1

56. (a) The area of the right triangle shown in Figure 2.5.18 is1

2r2 sin

π

ncos

π

n. Since there are

2n such right triangles, the area of the polygon is:

A(n) = 2n

�1

2r2 sin

π

ncos

π

n

�= nr2

�1

2sin

2π

n

�=

n

2r2 sin

2π

n

(b) A(100) ≈ 3.1395r2; A(1000) ≈ 3.1416r2

(c) Letting x = 2π/n (while noting that n = 2π/x) and substituting into A(n) above, weobtain:

A(n) =π

xr2 sinx = πr2

�sinx

x

�

From (10) of Section 2.4, we see that:

limn→∞

A(n) = πr2�limx→0

sinx

x

�= πr2

57. (a) limx→±∞

[f(x)− g(x)] = limx→±∞

�x2

x+ 1− (x− 1)

�

= limx→±∞

�x2

x+ 1− (x− 1)(x+ 1)

x+ 1

�= lim

x→±∞

x2 − (x2 − 1)

x+ 1

= limx→±∞

1

x+ 1= 0

(b) The graphs of f and g get closer and closer to each other when |x| is large.(c) g is a slant asymptote to f .


58. All points P are of the form (x, x2 + 1) while all points Q are of the form (x, x2). When they coordinates of P and Q are the same, we have x2

P + 1 = x2Q or xQ =

�x2P + 1, and thus

the horizontal distance between P and Q is |xQ − xP | = |�

x2P + 1− xP |. Thus:

limx→∞

|�x2 + 1− x| = lim

x→∞

��

��x2 + 1− x

��√x2 + 1 + x√x2 + 1 + x

��

= limx→∞

��x2 + 1− x2

√x2 + 1 + x

�� = limx→∞

��1√

x2 + 1 + x

�� = 0.

2.6 Limits — A Formal Approach

1. |10− 10| = 0 < � for any choice of δ.

2. |π − π| = 0 < � for any choice of δ.

3. |x− 3| < � whenever 0 < |x− 3| < �. Choose δ = �.

4. |2x− 8| = 2|x− 4| < � whenever 0 < |x− 4| < �/2. Choose δ = �/2.

5. |x+ 6− 5| = |x+ 1| < � whenever 0 < |x− (−1)| < �. Choose δ = �.

6. |x− 4− (−4)| = |x− 0| < � whenever 0 < |x− 0| < �. Choose δ = �.

7. |3x+ 7− 7| = 3|x− 0| < � whenever 0 < |x− 0| < �/3. Choose δ = �/3.

8. |9− 6x− 3| = |6− 6x| = 6|x− 1| < � whenever 0 < |x− 1| < �/6. Choose δ = �/6.

9.

��2x− 3

4− 1

4

�� =1

4|2x− 4| = 1

2|x− 2| < � whenever 0 < |x− 2| < 2�. Choose δ = 2�.

10. |8(2x+ 5)− 48| = |16x− 8| = 16

��x− 1

2

�� < � whenever 0 < |x− 2| < �/16. Chose δ = �/16.

11.

��x2 − 25

x+ 5− (−10)

�� = |x−5+10| = |x− (−5)| < � whenever 0 < |x− (−5)| < �. Choose δ = �.

12.

��x2 − 7x+ 12

2x− 6−�1

2

�� =

��(x− 3)(x− 4)

2(x− 3)+

1

2

�� =1

2|x − 4 + 1| =

1

2|x − 3| < � whenever

0 < |x− 3| < 2�. Choose δ = 2�.

13.

��8x5 + 12x4

x4− 12

�� = |8x+12−12| = 8|x−0| < � whenever 0 < |x−0| < �/8. Choose δ = �/8.

14.

��2x3 + 5x2 − 2x− 5

x2 − 1− 7

�� =��(2x+ 5)(x2 − 1)

x2 − 1− 7

�� = |2x + 5 − 7| = |2x − 2| = 2|x − 1| < �

whenever 0 < |x− 1| < �/2. Choose δ = �/2.

15. |x2 − 0| = |x− 0|2 < � whenever 0 < |x− 0| < √�. Choose δ =

√�.

16. |8x3 − 0| = 8|x− 0|3 < � whenever 0 < |x− 0| < 3√�/2. Choose δ = 3

√�/2.

2.6. LIMITS — A FORMAL APPROACH 99

17. |√5x− 0| =

√5|x− 0|1/2 < � whenever 0 < x < �2/5. Choose δ = �2/5.

18. |√2x− 1− 0| =

√2|x− 1/2|1/2 < � whenever 1/2 < x < 1/2 + �2/2. Choose δ = �2/2.

19. |2x− 1− (−1)| = |2x| = 2|x− 0| < � whenever 0− �/2 < x < 0. Choose δ = �/2.

20. |3− 3| = 0 < � whenever x > 1, for any choice of δ.

21. Note that |x2 − 9| = |x− 3||x+ 3| and consider only values of x for which |x− 3| < 1. Then2 < x < 4 and 5 < x + 3 < 7, so |x + 3| < 7. Thus, |x2 − 9| = |x − 3||x + 3| < 7|x − 3| < �whenever |x− 3| < �/7. Choose δ = min{1, �/7}.

22. Note that |2x2 + 4− 12| = 2|x2 − 4| = 2|x− 2||x+ 2| and consider only values of x for which|x − 2| < 1. Then 1 < x < 3 and 3 < x + 2 < 5, so |x + 2| < 5. Thus |2x2 + 4 − 12| =2|x− 2||x+ 2| < 10|x− 2| < � whenever |x− 2| < �/10. Choose δ = min{1, �/10}.

23. Note that |x2 − 2x+ 4− 3| = |x− 1|2 < � whenever |x− 1| < √�. Choose δ =

√�.

24. Note that |x2+2x−35| = |x−5||x+7| and consider only values of x for which |x−5| < 1. Then4 < x < 6 and 11 < x+7 < 13, so |x+7| < 13. Thus |x2+2x−35| = |x−5||x+7| < 13|x−5|�whenever |x− 5| < �/13. Choose δ = min{1, �/13}.

25. We need to show |√x−√a| < � whenever 0 < |x− a| < δ for an appropriate choice of δ. For

δ =√a�, we have

|√x−

√a| = |

√x−

√a| ·

√x+

√a√

x+√a=

|x− a|√x+

√a<

|x− a|√a

<

√a�√a

= �

whenever 0 < |x− a| < δ. Thus, limx→a

√x =

√a.

26. We need to show that |1/x− 1/2| < �, whenever 0 < |x− 2| < δ, for an appropriate choice ofδ. Without loss of generality, we may assume that δ < 1. Then |x− 2| < 1 or 1 < x < 3. Forthese values of x, 1/3 < 1/x < 1. Then, for δ = 2�, we have

��1

x− 1

2

�� =1

2

�1

x

�|2− x| < 1

2(1)|x− 2| < 1

2(2�) = �

whenever 0 < |x− 2| < δ. Thus, limx→2

1/x = 1/2.

27. Assume limx→1

f(x) = L. Take � = 1. Then there exists δ > 0 such that |f(x)−L| < 1 whenever

0 < |x− 1| < δ. To the right of 1, choose x = 1 + δ/2.

Since 0 < |1 + δ/2− 1| = |δ/2| < δ,

we must have |f(1 + δ/2)− L| = |0− L| = |L| < 1,

or −1 < L < 1.

To the left of 1, choose x = 1− δ/2.

Since 0 < |1− δ/2− 1| = |− δ/2| < δ,

we must have |f(1− δ/2)− L| = |2− L| < 1,

or 1 < L < 3.


Since no L can satisfy the conditions that −1 < L < 1 and 1 < L < 3, we conclude thatlimx→1

f(x) does not exist.

28. Assume limx→3


0 < |x− 3| < δ. To the right of 3, choose x = 3 + δ/2.

Since 0 < |3 + δ/2− 3| = |δ/2| < δ,

we must have |f(3 + δ/2)− L| = |− 1− L| = |L+ 1| < 1,

or −2 < L < 0.

To the left of 3, choose x = 3− δ/2.

Since 0 < |3− δ/2− 3| = |− δ/2| < δ,

we must have |f(3− δ/2)− L| = |1− L| = |L− 1| < 1,

or 0 < L < 2.

Since no L can satisfy the conditions that −2 < L < 0 and 0 < L < 2, we conclude thatlimx→3


29. Assume limx→0


0 < |x− 0| < δ. To the right of 0, choose x = δ/2.

Since 0 < |δ/2− 0| = |δ/2| < δ,

we must have |f(δ/2)− L| = |2− δ/2− L| < 1,

or 1− δ/2 < L < 3− δ/2.

To the left of 0, choose x = −δ/2.

Since 0 < |− δ/2− 0| = |− δ/2| < δ,

we must have |f(−δ/2)− L| = |− δ/2− L| < 1,

or −1− δ/2 < L < 1− δ/2.

Since no L can satisfy the conditions that 1− δ/2 < L < 3− δ/2 and −1− δ/2 < L < 1− δ/2,we conclude that lim

x→0f(x) does not exist.

30. Assume limx→0


0 < |x − 0| < δ. Since |f(x) − L| < 1 for all x such that 0 < |x| < δ, we may assume thatδ < 2. To the right of 0, choose x = δ/2.

Since 0 < |δ/2− 0| = |δ/2| < δ,

we must have |f(δ/2)− L| = |2/δ − L| = |L− 2δ| < 1,

or 2/δ − 1 < L < δ/2 + 1.

To the left of 0, choose x = −δ/2.

Since 0 < |− δ/2− 0| = |δ/2| < δ,

we must have |f(−δ/2)− L| = |− 2/δ − L| = |L+ 2/δ| < 1,

or −2/δ − 1 < L < −2/δ + 1.

2.6. LIMITS — A FORMAL APPROACH 101

Since we assumed δ < 2, we have

1 < 2/δ or 0 < 2/δ − 1

and −1 > −2/δ or 0 > −2/δ + 1.

Having established 2/δ − 1 < L < δ/2 + 1 and −2/δ − 1 < L < −2/δ + 1, these imply

0 < L < 2δ + 1 and −2/δ − 1 < L < 0.

Since it is impossible for L to satisfy both of these inequalities, limx→0


31. By Definition 2.6.5(i), for any � > 0 we must find an N > 0 such that��5x− 1

2x+ 1− 5

2

�� < � whenever x > N.

Now by considering x > 0,��5x− 1

2x+ 1− 5

2

�� =��

−7

4x+ 2

�� =7

4x+ 2<

7

4x< �

whenever x > 7/4�. Hence, choose N = 7/4�.

32. By Definition 2.6.5(i), for any � > 0 we must find an N > 0 such that��

2x

3x+ 8− 2

3

�� < � whenever x < N.

Now by considering x > 0,��

2x

3x+ 8− 2

3

�� =��

−16

9x+ 24

�� =16

9x+ 24<

16

9x< �

whenever x > 16/9�. Hence, choose N = 16/9�.

33. By Definition 2.6.5(ii), for any � > 0 we must find an N < 0 such that��10x

x− 3− 10


Now by considering x < 0,��10x

x− 3− 10

�� =��

30

x− 3

�� =��

30

−(−x+ 3)

�� =30

−x+ 3<

30

x< �

whenever x < −30/�. Hence, choose N = −30/�.

34. By Definition 2.6.5(ii), for any � > 0 we must find an N < 0 such that��

x2

x2 + 3− 1


Now by considering x < 0,��

x2

x2 + 3− 1

�� =��

−3

x2 + 3

�� =3

x2 + 3<

3

x2< �

whenever x2 > 3/� or x < −�3/�. Hence, choose N = −

�3/�.


35. We need to show |f(x) − 0| = |f(x)| < � whenever 0 < |x − 0| = |x| < δ for an appropriatechoice of δ. For δ = �,

|f(x)| =�|x|, x rational

0, x irrational< � whenever 0 < |x| < δ.

Thus, limx→0

f(x) = 0.

2.7 The Tangent Line Problem

1.-10 -5 5 10

-10

10

change in x = h = 2.5− 2 = 0.5

change in y = f(2 + 0.5)− f(2) = 2.75− 5 = −2.25

msec =change in y

change in x=

−2.25

0.5= −4.5

2.-10 -5 5 10

-10

10

change in x = h = 0− (−1/4) = 1/4

change in y = f(0 + 1/4)− f(0) = 17/16− 0 = 17/16

msec =change in y

change in x=

17/16

1/4=

17

4

3.-10 -5 5 10

-10

10

change in x = h = −1− (−2) = 1

change in y = f(−2 + 1)− f(−2) = −1− (−8) = 7

msec =change in y

change in x=

7

1= 7

4.-5 5

-5

5

change in x = h = 1− 0.9 = 0.1

change in y = f(1 + 0.1)− f(1) = 10/11− 1 = −1/11

msec =change in y

change in x=

−1/11

1/10= −10

11

2.7. THE TANGENT LINE PROBLEM 103

5.

-3

3

!!

2

change in x = h =2π

3− π

2=

π

6

change in y = f�π2+

π

6

�− f

�π2

�= sin

2

3π − 1 =

√3/2− 1

msec =change in y

change in x=

√3/2− 1

π/6=

3√3− 6

π

6.

-3

3

–! –!

3

change in x = h = −π

3−�−π

2

�=

π

6

change in y = f�−π

3+

π

6

�− f

�−π

3

�= cos

�−π

6

�− 1

2

=

√3

2− 1

2=

√3− 1

2

msec =change in y

change in x=

(√3− 1)/2

π/6=

3√3− 3

π

7. f(a) = f(3) = 3; f(a+ h) = f(3 + h) = (h+ 3)2 − 6

f(a+ h)− f(a) = [(h+ 3)2 − 6]− 3 = [(h2 + 6h+ 9)− 6]− 3 = h2 + 6h = h(h+ 6)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(h+ 6)

h= lim

h→0(h+ 6) = 6

With point of tangency (3, 3), we have y − 3 = 6(x− 3) or y = 6x− 15.

8. f(a) = f(−1) = 7; f(a+ h) = f(−1 + h) = −3(h− 1)2 + 10

f(a+ h)− f(a) = [−3(h− 1)2 + 10]− 7 = [(−3h2 + 6h− 3) + 10]− 7 = −3h2 + 6h = h(6− 3h)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(6− 3h)

h= lim

h→0(6− 3h) = 6

With point of tangency (−1, 7), we have y − 7 = 6(x+ 1) or y = 6x+ 13.

9. f(a) = f(1) = −2; f(a+ h) = f(1 + h) = (h+ 1)2 − 3(h+ 1)

f(a+ h)− f(a) = [(h+ 1)2 − 3(h+ 1)]− (−2) = (h2 − h− 2)− (−2) = h2 − h = h(h− 1)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(h− 1)

h= lim

h→0(h− 1) = −1

With point of tangency (1,−2), we have y + 2 = −(x− 1) or y = −x− 1.

10. f(a) = f(−2) = −17; f(a+ h) = f(−2 + h) = −(h− 2)2 + 5(h− 2)− 3

f(a+ h)− f(a) = [−(h− 2)2 + 5(h− 2)− 3]− (−17)

= (−h2 + 9h− 17)− (−17) = −h2 + 9h = h(9− h)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(9− h)

h= lim

h→0(9− h) = 9

With point of tangency (−2,−17), we have y + 17 = 9(x+ 2) or y = 9x+ 1.


11. f(a) = f(2) = −14; f(a+ h) = f(2 + h) = −2(h+ 2)3 + (h+ 2)

f(a+ h)− f(a) = [−2(h+ 2)3 + (h+ 2)]− (−14)

= (−2h3 − 12h2 − 23h− 14)− (−14) = h(−2h2 − 12h− 23)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(−2h2 − 12h− 23)

h

= limh→0

(−2h2 − 12h− 23) = −23

With point of tangency (2,−14), we have y + 14 = −23(x− 2) or y = −23x+ 32.

12. f(a) = f(1/2) = −3; f(a+ h) = f(1/2 + h) = 8(h+ 1/2)3 − 4

f(a+ h)− f(a) = [8(h+ 1/2)3 − 4]− (−3)

= (8h3 + 12h2 + 6h− 3)− (−3) = 2h(4h2 + 6h+ 3)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

2h(4h2 + 6h+ 3)

h

= limh→0

2(4h2 + 6h+ 3) = 6

With point of tangency (1/2,−3), we have y + 3 = 6(x− 1/2) or y = 6x− 6.

13. f(a) = f(−1) = −1/2; f(a+ h) = f(−1 + h) =1

2(h− 1)

f(a+ h)− f(a) =1

2(h− 1)−

�−1

2

�=

1 + h− 1

2(h− 1)=

h

2(h− 1)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

�h

2(h− 1)· 1h

�

= limh→0

1

2(h− 1)= −1

2

With point of tangency (−1,−1/2), we have y +1

2= −1

2(x+ 1) or y = −x

2.

14. f(a) = f(2) = 4; f(a+ h) = f(2 + h) =4

(h+ 2)− 1

f(a+ h)− f(a) =4

(h+ 2)− 1− 4 =

4− 4h− 4

h+ 1=

−4h

h+ 1

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

�−4h

h+ 1· 1h

�= lim

h→0

−4

h+ 1= −4

With point of tangency (2, 4), we have y − 4 = −4(x− 2) or y = −4x+ 12.

15. f(a) = f(0) = 1; f(a+ h) = f(h) =1

(h− 1)2

f(a+ h)− f(a) =1

(h− 1)2− 1 =

−h2 + 2h

(h− 1)2=

h(2− h)

(h− 1)2

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

�h(2− h)

(h− 1)2· 1h

�= lim

h→0

2− h

(h− 1)2= 2

With point of tangency (0, 1), we have y − 1 = 2(x− 0) or y = 2x+ 1.


16. f(a) = f(−1) = 12; f(a+ h) = f(−1 + h) = 4− 8

−1 + h

f(a+ h)− f(a) =

�4− 8

−1 + h

�− 12 =

−8

−1 + h− 8 =

−8− 8h+ 8

h− 1=

−8h

h− 1

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

�−8h

h− 1· 1h

�= lim

h→0

−8

h− 1= 8

With point of tangency (−1, 12), we have y − 12 = 8(x+ 1) or y = 8x+ 20.

17. f(a) = f(4) = 2; f(a+ h) = f(4 + h) =√4 + h

f(a+ h)− f(a) =√4 + h− 2 = (

√4 + h− 2)

√4 + h+ 2√4 + h+ 2

=4 + h− 4√4 + h+ 2

=h√

4 + h+ 2

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

�h√

4 + h+ 2· 1h

�

= limh→0

1√4 + h+ 2

=1

4

With point of tangency (4, 2), we have y − 2 =1

4(x− 4) or y =

1

4x+ 1.

18. f(a) = f(1) = 1; f(a+ h) = f(1 + h) =1√h+ 1

f(a+ h)− f(a) =1√h+ 1

− 1 =1−

√h+ 1√

h+ 1=

1−√h+ 1√

h+ 1· 1 +

√h+ 1

1 +√h+ 1

=1− h− 1√h+ 1 + h+ 1

=−h√

h+ 1 + h+ 1

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

�−h√

h+ 1 + h+ 1· 1h

�

= limh→0

−1√h+ 1 + h+ 1

= −1

2

With point of tangency (1, 1), we have y − 1 = −1

2(x− 1) or y = −1

2x+

3

2.

19. f(a) = f(π/6) = 1/2; f(a+ h) = f(π/6 + h) = sin(π/6 + h)

f(a+ h)− f(a) = sin�π6+ h

�− 1

2= sin

π

6cosh+ cos

π

6sinh− 1

2

=1

2cosh+

√3

2sinh− 1

2=

1

2(cosh− 1) +

√3

2sinh

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

�1

2· cosh− 1

h+

√3

2· sinh

h

�

= (1/2)(0) + (√3/2)(1) =

√3/2

With point of tangency (π/6, 1/2), we have y − 1

2=

√3

2

�x− π

6

�or y =

√3

2x−

√3π

12+

1

2.


20. f(a) = f(π/4) =√2/2; f(a+ h) = f(π/4 + h) = cos(π/4 + h)

f(a+ h)− f(a) = cos�π4+ h

�−

√2

2= cos

π

4cosh− sin

π

4sinh−

√2

2

=

√2

2cosh−

√2

2sinh−

√2

2=

√2

2(cosh− sinh− 1)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

√2

2

�cosh− 1

h− sinh

h

�

= (√2/2)(0− 1) = −

√2/2

With point of tangency (π/4,√2/2), we have y−

√2

2= −

√2

2

�x− π

4

�or y = −

√2

2x+

√2π

8+

√2

2.

21. f(a) = f(1) = 1; f(a+ h) = f(1 + h) = (h+ 1)2

f(a+ h)− f(a) = [(h+ 1)2]− 1 = (h2 + 2h+ 1)− 1 = h(h+ 2)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(h+ 2)

h= lim

h→0(h+ 2) = 2

The slope of the tangent at the blue point (1, 1) is 2. The slope of the line through (1, 1)and (4, 6) is m = (6 − 1)/(4 − 1) = 5/3. Since the slopes are not equal, then this line is nottangent to the graph.

22. Since there is more than one line, we first find the slope of the tangent line at the “generalpoint” (a, f(a)).

f(a) = a2; f(a+ h) = (h+ a)2

f(a+ h)− f(a) = [(h+ a)2]− (a2) = (h2 + 2ha+ a2)− a2 = h(h+ 2a)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(h+ 2a)

h= lim

h→0(h+ 2a) = 2a

Now that we have determined that mtan = 2a, then the slope of the tangent at the bluepoint (−1, 1) is mtan(−1) = 2(−1) = −2. The slope of the line through (−1, 1) and (1,−3) ism = (−3− 1)/(1+1) = −2. Since the slopes are equal, then this line is tangent to the graph.

The slope of the tangent at the blue point (3, 9) is mtan(3) = 2(3) = 6. The slope of the linethrough (3, 9) and (1,−3) is m = (9 + 3)/(3 − 1) = 6. Since the slopes are equal, then thisline is tangent to the graph.

23. We know that the points (2, 0) and (6, 4) are on the tangent line, so its equation is

y − 0 =0− 4

2− 6(x− 2) or y = x− 2

The line’s y-intercept is (0,−2).

24. We know that the points (0, 4) and (7, 0) are on the tangent line, so its equation is

y − 0 =4− 0

0− 7(s− 7) or y = −4

7x+ 4


Since the point of tangency (−5, f(−5)) is on this tangent line, then

f(−5) = −4

7(−5) + 4 =

48

7

25. f(a) = −a2 + 6a+ 1; f(a+ h) = −(h+ a)2 + 6(h+ a) + 1

f(a+ h)− f(a) = [−(h+ a)2 + 6(h+ a) + 1]− (−a2 + 6a+ 1)

= −h2 − 2ha− a2 + 6h+ 6a+ 1− (−a2)− 6a− 1

= −h2 − 2ha+ 6h = h(−h− 2a+ 6)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(−h− 2a+ 6)

h= lim

h→0(−h− 2a+ 6) = −2a+ 6

The tangent line is horizontal when mtan = 0, so we substitute and solve mtan = 0 = −2a+6,yielding 2a = 6 and a = 3. Thus, the tangent line is horizontal at (3, f(3)) = (3, 10).

26. f(a) = 2a2 + 24a− 22; f(a+ h) = 2(h+ a)2 + 24(h+ a)− 22

f(a+ h)− f(a) = [2(h+ a)2 + 24(h+ a)− 22]− (2a2 + 24a− 22)

= 2h2 + 4ha+ 2a2 + 24h+ 24a− 22− 2a2 − 24a− (−22)

= 2h2 + 4ha+ 24h = h(2h+ 4a+ 24)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(2h+ 4a+ 24)

h= lim

h→0(2h+ 4a+ 24) = 4a+ 24

The tangent line is horizontal when mtan = 0, so we substitute and solve mtan = 0 = 4a+24,yielding 4a = −24 and a = −6. Thus, the tangent line is horizontal at (−6, f(−6)) =(−6,−94).

27. f(a) = a3 − 3a; f(a+ h) = (h+ a)3 − 3(h+ a)

f(a+ h)− f(a) = [(h+ a)3 − 3(h+ a)]− (a3 − 3a)

= h3 + 3h2a+ 3ha2 + a3 − 3h− 3a− a3 − (−3a)

= h3 + 3h2a+ 3ha2 − 3h = h(h2 + 3ah+ 3a2 − 3)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(h2 + 3ah+ 3a2 − 3)

h

= limh→0

(h2 + 3ah+ 3a2 − 3) = 3a2 − 3

The tangent line is horizontal when mtan = 0, so we substitute and solve mtan = 0 = 3a2 − 3,yielding 3a2 = 3 and a = ±1. Thus, the tangent line is horizontal at (−1, f(−1)) = (−1, 2)and (1, f(1)) = (1,−2).


28. f(a) = −a3 + a2; f(a+ h) = −(h+ a)3 + (h+ a)2

f(a+ h)− f(a) = [−(h+ a)3 + (h+ a)2]− (−a3 + a2)

= −h3 − 3h2a− 3ha2 − a3 + h2 + 2ah+ a2 − (−a3)− a2

= −h3 − 3h2a− 3ha2 + h2 + 2ah = h(−h2 − 3ah− 3a2 + h+ 2a)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(−h2 − 3ah− 3a2 + h+ 2a)

h

= limh→0

(−h2 − 3ah− 3a2 + h+ 2a) = −3a2 + 2a

The tangent line is horizontal whenmtan = 0, so we substitute and solvemtan = 0 = −3a2+2a,yielding a(3a−2) = 0 and a = 0, 2/3. Thus, the tangent line is horizontal at (0, f(0)) = (0, 0)and (2/3, f(2/3)) = (2/3, 4/27).

29. vave =change of distance

change in time=

290 mi

5 h= 58 mi/h


change in time=

1/2 mi

40 s=

(1/2 mi)

(40 s)/(3600 s/h)=

1/2 mi

1/90 h= 45 mi/h

The car will not be stopped for speeding.


change in time; 920 km/h =

3500 km

t; t ≈ 3.8 h = 3 h 48 min


change in time=

20 mi− 10 mi

3 16 h− 1 1

2 h=

20 mi− 10 mi

19/6 h− 3/2 h=

10 mi

5/3 h= 6 mi/h

33. ∆s = s(t0+∆t)−s(t0) = f(3+∆t)−f(3) = [−4(3+∆t)2+10(3+∆t)+6]−0 = −14∆t−4∆t2

The instantaneous velocity at t = 3 is

v(3) = lim∆t→0

∆s

∆t= lim

∆t→0

−14∆t− 4∆t2

∆t= lim

∆t→0(−14− 4∆t) = −14.

34. ∆s = s(t0 +∆t)− s(t0) = f(∆t)− f(0) = ∆t2 +1

5∆t+ 1− 1 =

5∆t3 +∆t2 − 5∆t

5∆t+ 1= −5

The instantaneous velocity at t = 0 is

v(0) = lim∆t→0

∆s

∆t= lim

∆t→0

5∆t3 +∆t2 − 5∆t

(5∆t+ 1)∆t= lim

∆t→0

5∆t2 +∆t− 5

5∆t+ 1= −5.

35. (a) ∆s = s(t0 +∆t)− s(t0) = f(1/2 +∆t)− f(1/2) = −4.9(1/2 +∆t)2 + 122.5− 121.275

= −4.9∆t2 − 4.9∆t

The instantaneous velocity at t = 1/2 is

v(1/2) = lim∆t→0

∆s

∆t= lim

∆t→0

−4.9∆t2 − 4.9∆t

∆t= lim

∆t→0(−4.9∆t− 4.9) = −4.9 m/s.


(b) The ball hits the ground when s(t) = 0:

−4.9t2 + 122.5 = 0; t2 = 122.5/4.9; t = 5 s.

(c) Since the ball impacts at t = 5,

∆s = s(t0 +∆t)− s(t0) = f(5 +∆t)− f(5) = [−4.9(5 +∆t)2 + 122.5]− [−4.9(5)2 + 122.5]

= −49∆t2 − 49∆t

The impact velocity at t = 5 is

v(5) = lim∆t→0

∆s

∆t= lim

∆t→0

−49∆t2 − 49∆t

∆t= lim

∆t→0(−49∆t− 49) = −49 m/s.

36. (a) Setting − 12gt

2 + h = 0 and solving for t > 0, we obtain t =�2h/g.

(b) Earth: timpact =�2(100)/32 = 2.5 s

Mars: timpact =�2(100)/12 ≈ 4.08 s

Moon: timpact =�2(100)/5.5 ≈ 6.03 s

(c) ∆s = s(t0 +∆t)− s(t0) = −1

2g(t0 +∆t)2 + h− (−1

2gt20 + h) = −1

2g∆t2 − gt0∆t

The instantaneous velocity at timpact is

v(timpact) = lim∆t→0

∆s

∆t= lim

∆t→0

−1

2g∆t2 − gt0∆t

∆t= lim

∆t→0

�−1

2g∆t− gt0

�= −gt0.

(d) The impact velocities are

vEarth = −(32)(2.5) = −80 ft/s

vMars ≈ −(12)(4.08) = −48.96 ft/s

vMoon ≈ −(5.5)(6.03) = −33.165 ft/s.

37. (a) s(t) = −16t2 + 256t

s(2) = −16(22) + 256(2) = 448 ft

s(6) = −16(62) + 256(6) = 960 ft

s(9) = −16(92) + 256(9) = 1008 ft

s(10) = −16(102) + 256(10) = 960 ft

(b) s(5) = −16(52) + 256(5) = 880 ft

s(2) = 448 ft [from (a)]

vave =change of distance

change in time=

880 ft− 448 ft

5 s− 2 s= 144 ft/s


(c) s(7) = −16(72) + 256(7) = 1008 ft

s(9) = 1008 ft [from (a)]

vave =change of distance

change in time=

1008 ft− 1008 ft

9 s− 7 s=

0

2= 0 ft/s

At t = 7 s, the projectile is at a height of 1008 ft on its way upward. After it reachesa maximum height, it begins to fall downward and, at t = 9 s, the height is once again1008 ft. Since distance upward is positive and distance downward is negative, the netdistance is zero.

(d) The projectile hits the ground when s(t) = 0:

−16t2 + 256t = 0; 16t2 = 256t; t = 256/16 = 16 s

(e) For some general time t:∆s = s(t+∆t)− s(t) = [−16(t+∆t)2 + 256(t+∆t)]− (−16t2 + 256t)

= −16∆t2 + 256∆t− 32t∆t = ∆t(−16∆t+ 256− 32t)The instantaneous velocity at a general time t is

v(t) = lim∆t→0

∆s

∆t= lim

∆t→0

∆t(−16∆t+ 256− 32t)

∆t= lim

∆t→0(−16∆t+ 256− 32t) = (256− 32t) ft/s.

(f) From (d), the projectile impacts at t = 16 s. From (e), v(t) = 256 − 32t so v(16) =256− 32(16) = −256 ft/s.

(g) The maximum height is reached when v(t) = 0: 256 − 32t = 0 gives us t = 8 s. Sinces(t) = −16t2 + 256t, we have s(8) = −16(82) + 256(8) = 1024 ft.

38. (a) s(4) ≈ 1.3 ft; s(6) ≈ 2.7 ft

(b) vave ≈s(6)− s(4)

6− 4=

2.7− 1.3

2= 0.7 ft/s

(c) The instantaneous velocity at t = 0 is the slope of the tangent line to the graph at t = 0.In this case, v0 ≈ 1 ft/s.

(d) t ≈ 3 s

(e) The velocity is decreasing where the slopes of the tangent lines are decreasing; in thiscase, for 0 < t < 3.

(f) The velocity is increasing where the slopes of the tangent lines are increasing; in thiscase, for 3 < t < 7.

39. The slopes m of a tangent line at (a, f(a)) and m� of a tangent line at (−a, f(−a)) are:

m = limh→0

f(a+ h)− f(a)

h; m� = lim

h�→0

f(−a+ h�)− f(−a)

h�

As defined in Section 1.2, an even function is a function which is symmetric with respect tothe y-axis: f(−x) = f(x) for all x. Since f is even, then f(−a) = f(a) and f(−a + h�) =f(−[−a+ h�]) = f(a− h�), resulting in:

m� = limh�→0

f(a− h�)− f(a)

h� = limh�→0

f(a+ [−h�])− f(a)

h�

CHAPTER 2 IN REVIEW 111

Without loss of generality, we apply the substitution h� = −h to obtain:

m� = limh�→0

f(a+ [−h�])− f(a)

h� = limh→0

f(a+ h)− f(a)

−h= −m

40. The slopes m of a tangent line at (a, f(a)) and m� of a tangent line at (−a, f(−a)) are:

m = limh→0

f(a+ h)− f(a)

h; m� = lim

h�→0

f(−a+ h�)− f(−a)

h�

As defined in Section 1.2, an odd function is a function which is symmetric with respect tothe origin: f(−x) = −f(x) for all x. Since f is odd, then f(−a) = −f(a) and f(−a+ h�) =−f(−[−a+ h�]) = −f(a− h�), resulting in:

m� = limh�→0

−f(a− h�)− [−f(a)]

h� = − limh�→0

−[f(a− h�) + f(a)]

h�

Without loss of generality, we apply the substitution h� = −h to obtain:

m� = limh�→0

−[f(a− h�) + f(a)]

h� = limh→0

−[f(a+ h)− f(a)]

−h= m

41. To show that the graph of f(x) = x2+ |x| does not possess a tangent line at (0, 0), we examine

limh→0

f(0 + h)− f(0)

h= lim

h→0

[(0 + h)2 + |0 + h|]− 0

h= lim

h→0

h2 + |h|h

From the definition of absolute value, we see that

limh→0+

h2 + |h|h

=h2 + h

h= h+ 1 = 1

whereas

limh→0−

h2 + |h|h

=h2 − h

h= h− 1 = −1

Since the right-hand and left-hand limits are not equal, we conclude that limh→0

f(0 + h)− f(0)

h=

limh→0

h2 + |h|h

does not exist, and that therefore f has no tangent line at (0, 0).

Chapter 2 in Review

A. True/False

1. True

2. False; limx→5+

√x− 5 = 0.

3. False; limx→0−

|x|x

= −1.


4. False; limx→∞

e2x−x2= 0.

5. False; limx→0+

�tan−1 1

x

�=

π

2.

6. True

7. True

8. False; let f(x) = 0.

9. False; consider f(x) =1

x2, g(x) =

1

x4, and a = 0.

10. False; consider f(x) =1

x2, g(x) =

1

tan2 x, and a = 0.

11. False; consider f(x) = −x.

12. True

13. True; since f(−1) < 0 and f(1) > 0.

14. False; consider f(x) = 1 and g(x) = x− 2.

15. True

16. False; consider f(x) =

�−1, x < 0

1, x > 0and a = 0.

17. False; consider f(x) =

�1, x ≤ 3

2, x > 3.

18. True; since limx→a

[(x− a)f(x)] = [ limx→a

(x− a)][ limx→a

f(x)] = 0 · f(a) = 0.

19. True

20. False; limx→5

f(x) = 4 = f(5).

21. False; since

√x

x+ 1is undefined for x < 0.

22. False; the slope m of the tangent line at (3, f(3)) is 1. There is not enough information todetermine the value of f(3).


B. Fill in the Blanks

1. 4

2. 1

3. -1/5

4. -1/2

5. 0

6. 3/5

7. ∞

8. 0

9. 1

10. 1/4

11. 3−

12. 4

13. −∞

14. 0+

15. −2

16. Dividing by x2 we have 1− x2

3≤ f(x)

x2≤ 1. Since lim

x→0

�1− x2

3

�= 1 = lim

x→01, by the Squeeze

Theorem we have limx→0

f(x)

x2= 1.

17. 10

18. 8

19. continuous

20. 2

21. 9

22. Since f(x) = x2 is continuous, limx→−5

f(g(x)) = f( limx→−5

g(x)) = f(−9) = (−9)2 = 81.


C. Exercises

1.-5 5

-5

5

2.-5 5

-5

5

3.

-5

5

4.-3 3

-3

3

5. (a), (e), (f), (h)

6. (b), (e), (h)

7. (c), (h)

8. (b), (c), (d), (e), (f), (i)

9. (b), (c), (d), (e), (f)

10. (a), (g), (j)

11.-5 5

-5

5

The function is continuous everywhere.

12.

3 6

3

6

The function is discontinuous at x = 2 and x = 4.

13. (−∞,−1), (−1, 0), (0, 1), and (1,∞)


14. [−2, 1) and (1, 2]

15. (−∞,−√5) and (

√5,∞)

16. (nπ, nπ + π) for n = 0, 1, 2, . . .

17. For f(x) to be continuous at the number 3, we must have f(3) = 3k + 1 = limx→3+

(2 − kx).

Thus, we must solve for k in the equation 3k + 1 = 2− 3k, resulting in k = 1/6. Therefore:

f(x) =

x

6+ 1, x ≤ 3

2− x

6, x > 3

18. For f(x) to be continuous everywhere, we must have f(1) = 5 = limx→1+

(ax + b) and f(3) =

3a + b = limx→3+

(3x − 8). Thus, we get two equations 5 = a + b and 1 = 3a + b. Solving for a

an b yields a = −2, b = 7. Therefore:

f(x) =

x+ 4, x ≤ 1

−2x+ 7, 1 < x ≤ 3

3x− 8, x > 3

19. f(a) = f(2) = 32; f(a+ h) = f(2 + h) = −3(h+ 2)2 + 16(h+ 2) + 12

f(a+ h)− f(a) = [−3(h+ 2)2 + 16(h+ 2) + 12]− 32

= −3h2 − 12h− 12 + 16h+ 32 + 12− 32 = −3h2 + 4h = h(−3h+ 4)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(−3h+ 4)

h= 4


20. f(a) = f(−1) = −2; f(a+ h) = f(−1 + h) = (h− 1)3 − (h− 1)2

f(a+ h)− f(a) = [(h− 1)3 − (h− 1)2]− (−2)

= h3 − 3h2 + 3h− 1− h2 + 2h− 1− (−2)

= h3 − 4h2 + 5h = h(h2 − 4h+ 5)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(h2 − 4h+ 5)

h= 5

With point of tangency (−1,−2), we have y + 2 = 5(x+ 1) or y = 5x+ 3.

21. f(a) = f(1/2) = −2; f(a+ h) = f(1/2 + h) =−1

2(h+ 1/2)2

f(a+ h)− f(a) =−1

2(h+ 1/2)2− (−2) =

−1

2h2 + 2h+ 1/2− (−2)

=−1 + 4h2 + 4h+ 1

2h2 + 2h+ 1/2=

4h(h+ 1)

2h2 + 2h+ 1/2

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

4h(h+ 1)

(2h2 + 2h+ 1/2)h= 8


With point of tangency (1/2,−2), we have y + 2 = 8(x− 1/2) or y = 8x− 6.

22. f(a) = f(4) = 12; f(a+ h) = f(4 + h) = (h+ 4) + 4√h+ 4

f(a+ h)− f(a) = [(h+ 4) + 4√h+ 4]− 12 = [(h− 8) + 4

√h+ 4] · (h− 8)− 4

√h+ 4

(h− 8)− 4√h+ 4

=(h− 8)2 − 16(h+ 4)

(h− 8)− 4√h+ 4

=h2 − 16h+ 64− 16h− 64

(h− 8)− 4√h+ 4

=h(h− 32)

(h− 8)− 4√h+ 4

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(h− 32)

[(h− 8)− 4√h+ 4]h

=−32

−16= 2


23. f(a) = f(1) = 2; f(a+ h) = f(1 + h) = −4(h+ 1)2 + 6(h+ 1)

f(a+ h)− f(a) = [−4(h+ 1)2 + 6(h+ 1)]− 2 = −4h2 − 8h− 4 + 6h+ 6− 2

= −4h2 − 2h = h(−4h− 2)

mtan = limh→0

f(a+ h)− f(a)

h= lim

h→0

h(−4h− 2)

h= −2

With point of tangency (1, 2), we have y − 2 = −2(x − 1) or y = −2x + 4. Thus, the linethat is perpendicular to this line would have a slope of 1/2 and also passes through (1, 2),resulting in the equation y − 2 = (x− 1)/2 or y = (x+ 3)/2.

24. |2x + 5 − 7| = |2x − 2| = 2|x − 1| < � whenever |x − 1| < �/2. Thus, we choose δ = �/2 andso δ = 0.005 when � = 0.01. Finding δ proves that lim

x→1(2x+ 5) = 7.

zill calculo 4e manual de solucionario c02(1)

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