Section 1.4 Continuity and One-Sided Limits

126. (ax)-*Olimx12-c?sx x_-*lOira1x-c2osx

1 + cosx 1 + cos x

= lim 1 -- COS2X x-*?x~(1 + cos x)

=

xli--m40sXin2 Zx

1

+

1

COS

X

Section 1.4 Continuity and One-Sided Limits 73

(b) From part (a),

1

- cos xz

~x -1~ 2

1-cosx

~l_xz 2

~

cosx

~

1--

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)xl-*i4m+ f(x) = 3 (b) lira f(x) = 3

(c) lim f(x) = 3 The function is continuous at x = 4 and is continuous

on

2.

(a)

x

-*li-2r+a

f(x)

=-2

(b) lim f(x)=-2

(c) .~li-*m_zf(x) =-2 The function is continuous at x = -2,

3. (a)x-l*i3+mf(x) = 0 (b) lim f(x) = 0

(c) limf(x) = 0 x-*3

The function is NOT continuous at x = 3.

4. (a)x-l*i-m 3+ f(x) = 3 (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).

5. (a)x-l*i2+mf(x) =-3 (b) lim f(x) = 3

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

The function is NOT continuous at x = 2.

6. (a) lim f(x) = 0 x-*-I+ (b) lim f(x)= 2

(c) Ji-*m_,f(x) does not exist. The function is NOT continuous at x = -1.

7.

lxim-*~g-=x

1 +

8

1 -1 8 + 8 16

8. lim 3 - 3 - 3 x-*5- x+5 5+5 10

9.

lxim-*~5x+x'~-5---2-5---=-

lim x-*5+x

1 +

5

-

1 10

10.

xli-m,Z+.2x2

- x ~-~

=

lxir-*aZ+

1 x+2

-

1 4

11. lim ~ dxoes 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 = lim x-9 .~-~9-(x - 9)(w/7 + 3)

= lim _ -1 1 x--~9- %/X + 3 6

13. limX = lim---x =-1

x-~O- X x->O- X

14.

limlX-lO x~O+ x - 10

_ lim x-lO-1 x->~o+x - 10

1 1

15. lira x+Ax

~-,0- ~

x

=

lim x-(x+Ax)

~-~o- x(x + Ax)

-~-1=

lim~->o-x(x-A+xAx)

.-~-1x

= ~li0m-x(x-+1 ~)

-1

1

x(x + 0) x2

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

17. limf(x) = lixm+2 5 -

x->3-

x->3- 2 2

24. xl-i+m2+~(2x - ~d) : 2(2) - 2 : 2

18.x-l>i2m +vxf"(x]x-=>2+lim(-x2

+4x-2] !

=

2

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

x~2- x~2-

limf x 2

19.xl~ilm+ f(xx~)l+ = lim(x+l) = 2 limf(x) = lim(x~ +1)= 2

x~l- x~l-

limf x 2

~. lxi~ml+ f(xx)~l+= lim(1-x) = 0

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

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

x-+3

lim(2-~-x~) = 2-@3) = 5 and Xl.-i~m3+x (2 - I-x]) = 2 - @4) = 6.

j~,x) - x21 _ 4 has discontinuities at x = -2 and x = 2 because f(-2) and f(2) are not defined. 28. f(x)- Xx2 +-- 1l

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

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

? 2010 Brooks/Cole, Cengage Learning

Section 1.4 Continuity and One-Sided Limits

Ix,

30. f(x) = 2,

x < 1 x = 1 has a discontinuity at

2x- 1, x > 1

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

31. g(x) = ",/49 - x2 is continuous on [-7, 7].

32. f(t) = 3 - xi/-~ - t2 is continuous on [-3, 3].

33.

lim

x-~0-

f(x)

=x-3+0+=

lim

f(x).fis

continuous

on

[-1,

4].

34. g(2)is not defined, g is continuous on [-1, 2).

35? f(x) =_6 has a nonremovable discontinuity at x = O.

x

36? f(x) =-- x3- h2as a nonremovable discontinuity at

X =2.

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

jr, x) - xx2-6_ 36

has a nonremovable discontinuity at x = -6 because lim f(x) does not exist, and has a removable

discontinuity at x = 6 because

limf(x)

x-->6

=xli-m>6x1-+

1 6

1,2

47. f(x) = (x +x2+)2(x- 5)

has a nonremovable discontinuity at x = 5 because l,~im-~5f (x )doesneoxt ist, and has aremovable discontinuity at x = -2 because

.l~_i>m_2--~t::(x~x-*=-2lxi1r-a5

-1 . 7

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

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

limf x lira 1 1

39.

f(x)

-

4-

x1 2 _

(2

1 -

x)(2

+

x)

has

nonremovable

x-~x + 2 3

discontinuities at x = +2 becausex~2 ()limf x and

49. f(x)- x + 7I

x+7

:i_,m_2f(x) do not exist. 40. f(x) : x2 +-1---i-s-c~ontinuous for all real x. 41. f(x) = 3x - cos x is continuous for all real x.

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

has a nonremovable discontinuity at x = -7 because :i_+m_vf(X) does not exist.

50. f(x) - Ix - 8 has a nonremovable discontinuity at x-8

x 8= because lxi-m~f8x()does notexist.

f(x) : {X,x2, xX >-< 11

43. f(x) - x2x _is xnot continuous at x = 0, 1. Because

has a possible discontinuity at x = 1.

x 1 X2 --X x-1

for x ~ 0, x = 0 is a removable

discontinuity, whereas x = 1 is a nonremovable

discontinuity.

44. f(x) = x2x_hlas nonremovable discontinuities at

1. f(1)= 1

limf(x) = limx = 1 ] 2. limf(x) = limx2 ~=limf(x) = 1

x-M+

x-~l+

J

a. y(-1)l=im(f)x

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

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

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

? 2010 Brooks/Cole, Cengage Learning

76 Chapter 1 Limits and Their Properties

52. f(x)= f-2x2x, + 3, xx11 has a possible discontinuity at x = 1. 1. f(1) = 12 -- 1

limf(x) = lim(-2x+3)= 1] 2. x--,q- x->l-~limf(x) = 1

xl--i>ml+ f(x)x~.=l+ limx2 = 1 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)=--2+1=2

2

2.

lim x-~2xl~i.2m+

f(x)=x-l~i2m-~(.2--x + f(xx-)-~=2+ lim (3 -

1) x)

= 2)limf x 1,-,2 ( ) does

not

exist.

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

f-2x, x < 2

54. f(x) = ix2 _4x+1, x > 2

has a possible discontinuity at x = 2. 1. f(2) = -2(2) = -4

lim f(x) = lim (-2x) = -4 xl-i-,m, 2+f(x) = lixm~.2(+x2 -4x + 1) = ~-3liJm~'f-(x~)2does not exist. Therefore,f has a nonremovable discontinuity at x = 2.

1

has possible discontinuities at x = -1, x = 1.

1. 1(-1)=-1

f(1) = 1

2..~_~_, ()=lim f x -1 3. f(-1)= :-f,x()

x-~llimf(x) = 1 lifm(1f)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

56. f(x) = csc-6-,

x-3 < 2

[2,

x 31 > 2

I= csc--6-, 1 < x < 5

[2,

x < lorx > 5

has possible discontinuities at x = 1, x = 5.

1. 1 = csc-- = 2 6

2. limf(x) = 2

5 = csc-- = 2 6

limf x 2

3, f(1) 1=.~mlf(x)

f(5) = xl-im>5f (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

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

59. f(x) = Ix - 8~ has nonremovable discontinuities at each integer k.

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

a(1) - 4 = 3

64. 1(1)= 3

Find

a

so

that

x

-l->im1+

(ax

+

5)

=

3

a(1) + 5 : 3

a = -2.

65. f(2) : 8 Findasothatxl->im2+ ax2 =8 ~ a = s@2~ = 2.

66. limg(x) = lim4Sinx _ 4

x->O-

x->O- X

x-l->iOm+ g(xx-)->O=+ lim(a-Zx) = a

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

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