Answers & Hints for Selected Exercises 665rk+l 3 max{x - k!l' rk} < rk+l < x. Then, 'Vn EN, x - ~ < rn < rn+l < x,
so {rn} is a strictly increasing sequence of rational numbers converging to x.
- By (b), ax-yay =ax, so ax-y =ax /aY since aY-:/:- 0 by Thm. 5.6.6.
5. By (d) and (e), (%t = (ab-^1 )X = ax(b-^1 )X = ax(bx)-1 = ~=.
7. Let f(x) =ax, 0 < a < 1. Then f(x) = l/g(x) where g(x) = (~t. Then
~ > 1 so by Thm. 5.6.11 g is continuous and strictly increasing on JR with range
(0, oo). Thus, f is continuous and strictly decreasing on JR with range (0, oo). - Lett< 0 and consider the power function f(x) = xt. Then -t > 0. By Thm.
5.6.14, the function g(x) = x-t is strictly increasing, positive, and continuous
on (O,oo) with x-+oo lim g(x) = +oo. Since f(x) = l/g(x), f is positive, strictly
decreasing, and continuous on (O,oo) with lim f(x) = +0.
X-+00
- By Thm. 5.6.17, X-+00 lim (1 + .!.)x X = e. Thus,
(i) by Thm. 4.4.19, lim (1 + x)^1 fx = lim (1 + ~t = e;
x-+O+ x-+oo
(ii) by Ex. 10, lim (1 + .!. r = e, so by Thm. 4.4.19, lim (1 + x)^1 fx = e.
x-+-oo x x-.o-
.". lim(l + x )^1 fx = e.
x-+O
- (a) f: (0, oo)----> JR, and is 1-1 and onto, by definition of inverse.
(b) f is continuous on (O,oo) by Cor. 5.5.3.
(c) The function g(x) = ax is strictly increasing if a > 1 and strictly
decreasing if 0 < a < 1, by Thms. 5.6.11 and 5.6.12. Thus, f = g-^1 has the
same properties by Cor. 5 .5.3.
(d) a^0 = 1::::} g(O) = 1::::} g-^1 (1) = 0::::} f(l) = 0.
(e) Suppose a > 1. Since loga x is strictly decreasing, x > 1 ::::} loga x >
loga 1 = 0 and 0 < x < 1 ::::} loga x < loga 1 = o.
(g) Note that ax> 0 and loga(ax) = f(g(x)) = J(f-^1 (x)) = x. - Let a, b > 0, a, b-:/:-1, and x > 0. Let y = logb x. Then bY = x, so
y loga b = loga x.
Chapter 6
EXERCISE SET 6.1x^3 - x~- (b) lim lim (x5 + xox + x^2 ) = 3x5.
x-+xo x - x 0 x-+xo
3'±1 _ xo+l 2 ( )
(d) lim x-1 x^0 -l = lim Xo - x
x-+xo x - xo x-+xo (x - xo)(x - l)(xo - 1)-2
(xo -1)^2.