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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