656 Appendix C • Answers & Hints for Selected Exercises
and 3 02 > 0 3 0 <Ix - xol < 02 => g(x) > VM. Then
0 <Ix - xol < min{o1,1h} => f(x)g(x) > VMVM = M.- Suppose f(x) ::::; g(x) "ix E N 01 (xo), and suppose lim f(x) = +oo.
x--+xo
Let M > 0. Then 3 02 > 0 3 0 <Ix - xol < 02 => f(x) > M. Then
0 < Ix - xol < min{o1, 02} => f(x)::::; g(x) and f(x) > M => g(x) > M. - (a) f(x) = x\, g(x) = -,12, Xo = 0 (b) f(x) = ~' g(x) = -~, Xo = 0
(c) f(x) = ~' g(x) = -~, Xo = 0 (d) f(x) = ~ + L, g(x) = -~, xo = 0
- (a) lim (x-l)
2
x->1 x = 0. ( x !:.^1 )^2 >^0 when x > 1 and when^0 < x < 1, so by
Thm. 4.4.3, lim (x!:. 1 ) 2 = +oo and lim (x!:. 1 ) 2 = +oo .. ·. lim (x!:.l)2 = +oo.
x-+1- x-+1+ x-+l
(b) + oo, + oo, + oo,
(c), (f), (g) -oo, +oo, does not exist (d), (h) + oo, - oo, does not exist
(e) lim : 2 -::_^31 = 0. "':x
2
~l > 0 when x > 3 and < 0 when 1 < x < 3, so
X-+3-
lim x^2 -l = -oo, lim x^2 -^1 2 1
x-+3- x- x-+3+ x-^3 = +oo, and x-+3 lim x x--^3 does not exist.EXERCISE SET 4.4-B- lim f(x) = +oo ¢:? V(f) is unbounded above and VM > 0, 3 N > 0 3
x-++oo
x > N => f(x) > M. - lim f(x) = -oo ¢:? V(f) is unbounded below and VM > 0, 3 N > 0 3
x-+-co
x < -N => f(x) < -M.
(^4) · ()1a 6x+5 3x-1 - 2 l, - 12x+10^7 <7 12x· 'T' .1.0 ma k e 12x 7< c:, tk12x>l. a e 7 °£'
i.e., x > l~E· Thus, take N = l~E· (b) To make^3 x
2
;;;~-^1 > M, take x >
max{4, 2m/3}. (c) To make^1 x--;; > M, take lxl > N, where N = max{2, M}.
(d) Take x < -(~ + 1). (e) Take x > M + 1. (f) Take x < -M.
- We know lim x^1 = - oo, so assume n;::: 3 and odd; say n = 2k + 1. Then
x--+-oo
lim xn = lim x^2 k+l = lim x^2 k · lim x^1 = +oo · -oo = -oo (in the
X--+-00 X-+-00 X--+- 00 X--+-00
sense of Table 4.1). - n even => lim xn +oo by Thm. 4.4.18 (b), so by Thm. 4.4.21(c),
X--+-00
X--+-00 lim --1;. X = 0.
nodd=> lim xn=-oobyThm.4.4.18(c),sobyThm.4.4.21(d), lim -1;.=0.
x--+-oo x--+- oo x - (a) Let Xn = nn, Yn = ~+2nn. Then Xn--+ + oo, Yn--+ +oo, but XnSinxn =
0 --+ 0, while Yn sin Yn = Yn --+ +oo. .". lim x sin x does not exist.
x--+-oo