Section 1.4 Continuity and One-Sided Limits 73

126. (a) lim1-c°sx _ lira1-cosx 1 + cosx x-*O x2 x-*O x2 1 + cos x 1 -- COS2X

= lim x-*°x~(1 + cos x) sinZx = lim 2 x--40 X

1 1 + COS X

(b) From part (a), 1 - cos ~x 1~ 1-cosx ~l_xz ~ cosx ~ 1-xz 2 2

1 x2 forx ~ O. 2

(c) cos(0.1) ~ 1 - ~(0.1)2 = 0.995 (d) cos(0.1) ~ 0.9950, which agrees with part (c). 127. The graphing utility was set in degree mode, instead ofradian mode.

Section 1.4 Continuity and One-Sided Limits 1.(a) limf(x) =3 +

5. (a) limf(x) =-3 +

(b) lira f(x) = 3

(b) lim f(x) = 3

(c) lim f(x) = 3

(c) limf(x) does not exist x-*2

The function is continuous at x = 4 and is continuous

The function is NOT continuous at x = 2.

x-*4

on

x-*2

6. (a) lim f(x) = 0 + x-*-I

2. (a) lira f(x) =-2 + x -* -2

(b) lim f(x)=-2 (c) .~li-*m_zf(x) =-2 The function is continuous at x = -2, 3. (a) limf(x) =0 +

(b) lim f(x)= 2 (c) Ji-*m_,f(x) does not exist. The function is NOT continuous at x = -1. 1 7. lim~ = x-*g-x + 8

1 - 1 8 + 8 16

x-*3

(b) lim f(x) = 0 (c) limf(x) = 0 x-*3

The function is NOT continuous at x = 3. 4. (a) lim f(x) = 3 + x-*-3

3 8. lim x-*5- x+5

3 5+5

3 10

x-5 = lim 1 - 1 9. lim~x’~------x-*5+ - 25 x-*5+x + 5 10 1 .22 - x = lira 10. lim +x x-,Z ~-~ x-*Z+ x+2

1 4

(b) lim f(x)= 3 (c))i_,m_3f(x) = 3 The function is NOT continuous at x = -3 because f(-3) = 4 J~m_3f(x).

x 11. lim ~ does not exist because x-*-3-~X2 -- 9

~Xdecreases without bound as x -~ -3-.

-9

© 2010 Brooks/Cole, Cengage Learning

74

Chapter 1 Limits and Their Properties

12. lim ~-3 _ lim~X-3 xi~+3 x->9- x-9 x->9- x-9 ~+3 x-9 = lim .~-~9-(x - 9)(w/7 + 3) 1 x--~9- %/X + 3

= lim _ 1 15. lira x+Ax

~

~-,0-

1 -

13. limX = lim---x =-1 x-~O- X

x->O- X

_ lim x-lO-1 14. limlX-lO x~O+ x - 10 x->~o+x - 10

6

1 1 x = lim x-(x+Ax) --= lim

~

~-~o- x(x + Ax)

-Ax

1 .--

~->o-x(x + Ax) ~x

= lim

-1

~0-x(x + ~) -1

1

x2

x(x + 0)

= 2x + 0+1 = 2x+l

x+2 5 17. limf(x) = lim x->3x->3- 2 2

24. lim+~ (2x - ~d) : 2(2) - 2 : 2

18.x->2 limf(x] = lim(-x2 +4x-2] =2 +vx " ! x->2+

25. lim(2 - [-x~)does not exist because x-+3

limf(x) = lim(x2 -4x+6) = 2

x~2-

x-+2

lim(2-~-x~) = 2-@3) = 5

x~2-

limf x 2

and lim (2 - I-x]) = 2 - @4) = 6.

19. limf(x) + + = lim(x+l) = 2 x~l

X.-~3+x

x~l

limf(x) = lim(x~ +1)= 2

x~l-

x~l-

limf x 2 1 j~,x) - x2 _ 4 ~. limf(x) + += lim(1-x) = 0 x~l

x~l

21. lira cot x does not exist because x~g lira cot x and lira cot x do not exist. 22. lira sec x does not exist because x~/~

lira see x and lira sec x do not exist. .~(~/~)+ x~(~/~)-

z3. ~im (~x~- 7) = ~(3/- 7 = ~ ([x~ = 3for3 ~ x < 4)

has discontinuities at x = -2 and x = 2 because f(-2) and f(2) are not defined. X2 -- 1 28. f(x)- x + l

has a discontinuity at x = -1 because f(-1) is not defined.

29. f(x)= ~.x__.~_~ + x 2 has discontinuities at each integer k because lim f(x) ~ lim f(x). + x--~k- x-~k

© 2010 Brooks/Cole, Cengage Learning

Section 1.4 Continuity and One-Sided Limits

x 1 () = x =l because f(1)= 2 ~limf x-~, x 1.

I

x-6 jr, x) - x2 _ 36 has a nonremovable discontinuity at x = -6 because lim f(x) does not exist, and has a removable discontinuity at x = 6 because 1 1 limf(x) = lim , x-->6 x->6x + 6 12

31. g(x) = ",/49 - x2 is continuous on [-7, 7]. 32. f(t) = 3 - xi/-~ - t2 is continuous on [-3, 3].

f(x) = (x +x+2 2)(x- 5)

47. 33. lim f(x) = 3 += lim f(x).fis continuous on [-1, 4]. x-~0-

x-+0

34. g(2)is not defined, g is continuous on [-1, 2). 35° f(x) =_6 has a nonremovable discontinuity at x = O. x

3 has a nonremovable discontinuity at 36° f(x) =-- x- 2

has a nonremovable discontinuity at x = 5 because limf (x )doesnot exist, and has aremovable ,~-~5 discontinuity at x = -2 because 1 lim t:(x~ = lira .~_>_2--~ : x-*-2x -5

-1 . 7

x-1 48. f(x) = (x + 2)(x1)

X =2.

37. f(x) = x2 - 9 is continuous for all real x.

has a nonremovable discontinuity at x = -2 because xl~m_2f(x) does not exist, and has a removable

38. f(x) = x2 - 2x + 1 is continuous for all real x.

discontinuity at x = 1 because

1 limf x lira 1 x-~x + 2 3 12_ 1 39. f(x) - 4- x (2 - x)(2 + x) has nonremovable 49. f(x)- x + 7I discontinuities at x = +2 becausex~2 ()limf x and x+7 has a nonremovable discontinuity at x = -7 because :i_,m_2f(x) do not exist. 1 is continuous for all real x. 40. f(x) : x2 +------~ 41. f(x) = 3x - cos x is continuous for all real x. 42.

:()x = cosm 2is continuous for all real x.

is x not continuous at x = 0, 1. Because 43. f(x) - x2x _ x

1

X2 --X x-1

for x ~ 0, x = 0 is a removable

discontinuity, whereas x = 1 is a nonremovable discontinuity.

:i_+m_vf(X) does not exist. 50. f(x) - Ix - 8 has a nonremovable discontinuity at x-8 x8 = because limf x-~8x()does notexist. f(x) : {X,x2, xX >-< 11 has a possible discontinuity at x = 1. 1. f(1)= 1 limf(x) = limx = 1 ] 2. limf(x) = limx2 ~limf(x) = =1 x-M+

x has nonremovable discontinuities at

44. f(x) = x2 _ l

x = land x = -1 becausex.--~llimf(x)and- , :i_>m_lf(x) do not exist.

x-~l+

J

a. y(-1)= ()x limf f is continuous at x = 1, therefore, f is continuous for all real x.

x is continuous for all real x. 45. f(x)= x2 +------~

© 2010 Brooks/Cole, Cengage Learning

76

Chapter 1 Limits and Their Properties

52. f(x)= f-2x + 3, xx11 x2, has a possible discontinuity at x = 1. 1. f(1) = 12 -- 1 limf(x) = lim(-2x+3)= 1] 2. x--,qx->l-~limf(x) = 1 2=1 limf(x) = + + limx x-->l

x~.l

3. f(1)= limf(x) x--~l f is continuous at x = 1, therefore, f is continuous for all real x. 53. f(x) = +1, x < 2 [3-x, x>2 has a possible discontinuity at x = 2. 1. f(2)=--+1=2 2 2 lim f(x)= lim(--x + 1) = 2)limf x x-~2-~.2 2. x-~2,-,2 ( ) does not exist. lim + f(x) = + lim (3 - x) 1 x~.2

x--~2

Therefore, f has a nonremovable discontinuity at x = 2.

f

-2x, x 2 has a possible discontinuity at x = 2. 1. f(2) = -2(2) = -4 lim f(x) = lim (-2x) = -4 does not exist. 2 -4x + 1) = ~limf(x) lim+f(x) = lim(x -3J~’-~2 +

x--,, 2

x~.2

Therefore,f has a nonremovable discontinuity at x = 2.
1

has possible discontinuities at x = -1, x = 1. 1. 1(-1)=-1 2..~_~_, ()=lim f x -1 lim f x() 3. f(-1)= :-,

f(1) = 1 x-~llimf(x) = 1 limf f(1)x= x--.,1 ()

fis continuous at x = +1, therefore,f is continuous for all real x.

© 2010 Brooks/Cole, Cengage Learning

Section 1.4 Continuity and One-Sided Limits 77

t

~x x-3 < 2 = csc-6-, 56. f(x) [2, x 31 > 2

I

15 limf () x

fis continuous at x = 1 and x = 5, therefore, f is continuous for all real x. 57. f(x) = csc 2x has nonremovable discontinuities at integer multiples of z/2.

63. f(1)= 3 Find a so that lim (ax - 4) = 3 a(1) - 4 = 3

58. f(x) = tan"" has nonremovable discontinuities at each 2 2k + 1, k is an integer. 59. f(x) = Ix - 8~ has nonremovable discontinuities at each integer k.

64. 1(1)= 3 Find a so that lim+ (ax + 5) = 3 x --> 1

a(1) + 5 : 3 a = -2.

60. f(x) = 5 - lx] has nonremovable discontinuities at each integer k. 61. lim f(x) = 0 + x~.O

lim f(x) = 0 fis not continuous at x = -2. 5O

65. f(2) : 8 2 =8 ~ a = s@ Findasothat limax + 2~ = 2. x->2

66. limg(x) = lim4Sinx _ 4 x->O-

x->O- X

limg(x) =+ lim(a-Zx) = a +

x-->O

x-->O

Let a = 4.

62. lira f(x) = 0 lim f(x) = 0 fis not continuous at x = -4. 20

-10

© 2010 Brooks/Cole, Cengage Learning

78

Chapter 1 Limits and Their Properties

67. Find a and b such that lira (ax+b) :-a+b : 2and lim(ax+b) : 3a+b =-2. + x-->-I

x--~3-

a - b = -2

(+)3a + b =-2 4a

= -4 a = -1

t;

x 3 ,

68. limg(x) = limx2 - a2 x-->a a = x-->a x->aX(--a) = 2a lim x +

--3x, 4 4 75. g(x) =X2x 5,x x> a = 4.

69.

IO

!

= (x -0

Continuous for all real x.

i~

8

/

1 70. f(g(x)) : air-~ _ 1 Nonremovable discontinuity at x = 1. Continuous for all x > 1. 1 1 71. f(g(x)) = (x2 + 5)- 6 = x2~-I

~, X_ 0 f(O) = 5(0) = 0 limf(x) = lim(C°sx-1) = 0

Nonremovable discontinuities at x = +1 72. f(g(x)) = sin x~ Continuous for all real x 73. y : Ix]-x

x--~ O-

x-#O-

X

limf(x) =+ lim(5x) = 0 +

x-.> O

x.-~ O

Therefore, limf(x)= 0 = f(0)andfis continuous on the entire real line. (x = 0 was the only possible discontinuity.)

Nonremovable discontinuity at each integer 0,5

-1,5

1 74. h(x) : (x + 1)(x- 2) Nonremovable discontinuities at x = -1 and x = 2. 2

X jtxl = x~ + x + 2

Continuous on (-m, m)

78. f(x) : x’,f-~ + 3 Continuous on [-3, ~X

79. f(x): sec-~-Continuous on: ..., (-6,-2),(-2, 2),(2, 6), (6, 10) ....

S(x)- x+l Continuous on (0, © 2010 Brooks/Cole, Cengage Learning

Section 1.4 Continuity and One-Sided Limits 79

87. f(x) = x3 + x-1

sin x 81..f|x} = ~ X

f(x) is continuous on [0, 1].

3

f(0) =-land f(1)= 1

-2

The graph appears to be continuous on the interval [-4, 4]. Because f(0) is not defined, you know thatfhas a discontinuity at x = 0. This discontinuity is removable so it does not show up on the graph. X3 -- 8

f(0) =-3 and f(1)= 3

14

-4

The graph appears to be continuous on the interval [-4, 4]. Because f(2) is not defined, you know that f has a discontinuity at x = 2. This discontinuity is removable so it does not show up on the graph.

: 7x -

88. f(x) = x~ +5x-3 f(x) is continuous on [0,1].

82. f(x) - x- 2

a3.

By the Intermediate Value Theorem, f(c) = 0 for at least one value of c between 0 and 1. Using a graphing utility to zoom in on the graph of f(x), you find that x ~ 0.68. Using the root feature, you find that x ~ 0.6823.

+ 4 is continuous on the interval

[1, 2]. f(1) = 37 and f(2) = -~.Bythe Intermediate Value Theorem, there exists a number c in [1, 2] such that f(c) = O. 84. f(x) = x3 + 5x - 3 is continuous on the interval [0, 1]. f(0) = -3 and f(1) = 3, By the Intermediate Value Theorem, there exists a number c in [0,1] such that

89. g(t) = 2cost-3t g is continuous on [0,1]. g(0) = 2 > 0and g(1)~ -1.9 < 0. By the Intermediate Value Theorem, g(c) = 0 for at least one value of c between 0 and 1. Using a graphing utility to zoom in on the graph of g(t), you find that t ~ 0.56. Using the root feature, you find that t = 0.5636. 90. h(O) = l+O-3tanO h is continuous on [0, 1].

f(c) = O. 85. f(x) : x: - 2 - cos x is continuous on [0, z]. f(0) = -3 and f(z) = n’: -1 ~ 8.87 > 0.Bythe Intermediate Value Theorem, f(c) = 0 for at least one value of c between 0 and z. 86. f(x) =--x+ tan

By the Intermediate Value Theorem, f(c) = 0 for at least one value of c between 0 and 1. Using a graphing utility to zoom in on the graph of f(x), you find that x ~ 0.56. Using the root feature, you find that x ~ 0.5641.

h(0) = 1 > 0 and h(1) ~ -2.67 < 0. By the Intermediate Value Theorem, h(e) = 0 for at least one value of c between 0 and 1. Using a graphing utility to zoom in on the graph of h(O), you find that 0 ~ 0.45. Using the root feature, you find that 0 ~ 0.4503.

is continuous on the interval [1, 4].

-5+ tan(~01 = -4.7 and f(4) = 4 + tan

~! 1.8. By the Intermediate

Value Theorem, there exists a number c in [1, 4] such that f(c) = O.

© 2010 Brooks/Cole, Cengage Learning