Answers & Hints for Selected Exercises 657(b) \Ix :f: 0, 0:::; j si~x j :::; - 11 x 1 --> 0, so by squeeze principle, x-+lim +oo .!. x sinx = 0.
(c) \Ix :f: 0, 0:::; J x .!. sin.!. x J :::; - 11 x 1 --> 0, so by squeeze, x--++oo lim .!. x sin.!. x = 0.- (a) Suppose L > 0. (=?) Suppose lim f(x) = +oo. Then f > 0 on a
x-oo
nbd of oo, and Xlim -+00 !(^1 X ) = 0. Using (4.4.23), Xlim --+00 -(g^1 X ) = X--+OO lim !(^1 X ). f(( 9 x)) X =
lim f(^1 x) · lim gf((:)) = 0 · L = 0. So, by Thm. 4.4.21, lim g(x) = +oo.
x--+oo x-+oo x--+oo
( ~) lim g(x) = +oo =? lim f(x) = lim (f((x)). g(x)) = lim f((x)).
x-+oo x--+oo x--+oo 9 x x--+oo 9 x
lim g(x) = L · +oo = +oo since L > 0 (Table 4.2 and Thm 4.4.23).
x-oo
(b) For L < 0, modify the proof of (a) appropriately. - (a) y = ~ (b) none (c), (d) y = 0 (e) none (f) y = -6
- (=?) Suppose lim f(x) = L. Let c > 0. Then 3 N E JR 3 x > N =?
x-oo
lf(x) - LI < c/2. Then x, y > N =? lf(x) - f(y)I :::; lf(x) - LI+ IL - f(y)I <
c/2 + c/2 = c.
( ~) Let c > 0. By hypothesis, 3 NE JR 3 x, y > N =? lf(x) - f(y)I < c.
Let {xn} be any sequence in V(f) converging to +oo. Then {f(xn)} is Cauchy
so it converges. Thus, \I sequences { Xn} in V(f) converging to +oo, { f (xn)}
converges. We must prove that all these sequences {f(xn)} have the same limit.
Let {xn}, {Yn} be sequences in V(f) converging to + oo. Then the sequence
{ x1, Y1, x2, Y2, X3, y3, · · ·} --> +oo, so the sequence {f (x1), f (Y1), f (x2), f (y2), · · ·}
converges. But then any two of its subsequences {f(xn)} and {f(yn)} must con-
verge to the same limit.
Chapter 5
EXERCISE SET 5.1- Suppose x 0 is an isolated point of V(f). Then
(a) :Jo> 0 3 N6(x 0 ) contains no point ofV(f). Let c > 0. Then \Ix E V(f),
Ix - xol < o =? x = xo =? lf(x) - f(xol = 0 < c, so f is continuous at xo.
(b) By Exercise 3.2.24, x 0 is not a cluster point ofV(f), so by Defn. 4.1.1,
lim f(x) does not exist.
X-+XQ
For f(x) = Jx3 - x^2 = Jx^2 (x -1), V(f) = {O} U [l,oo). Since 0 is an
isolated point of V(f), f is continuous at 0 but lim f(x) does not exist.
x-o
- For f(x) = Jx3 + 2x^2 + x = Jx(x - 1)^2 , V(f) = {-1} U [O, oo). Since -1 is
an iso lated point of V(f), f is continuous at -1 but lim f(x) does not exist.
X--+-1 - Let c > 0. Choose o = min{l, c/15}. Then
Ix - 21 < o =? Ix - 21 < 1 and Ix - 21 < c/ 15
=? 1 < x < 3 and Ix - 21 < c/ 15
=? 7 < 4x + 3 < 15 and Ix - 21 < c/ 15