288 Chapter 5 11 Continuous Functions
5.6.14 and 5.6.15; moreover, by Lemma 5.6.19, lim g(t) = e. By Theorem 5.1.14,
t-+O
lim(l + tx)^1 1t = lim [ (1 + tx)l/txt = lim f (g(t)) = f (lim g(t)) = f (e) = ex.
t-+O t-+O t-+O t-+O
Corollary 5.6.21 (Sequential Limit for e"') 't/x E JR, n--.ex> lim (1 + E)n n =ex.
Proof. Exercise 12. •
LOGARITHM FUNCTIONS
Definition 5.6.22 Let a> 0, a'/-1. We define the function f(x) = loga x to
be the inverse of the function g(x) =ax, which we have shown to be a 1-1,
onto, continuous function from JR to (0, +oo ). That is, y = loga x {::} aY = x.
The following two theorems show that this function satisfies a ll the prop-
erties that we expect logarithms to have.
Theorem 5.6.23 (Properties of Logarithms, I) Let a > 0, a '/- 1. The
function f ( x) = loga x has the following properties:
(a) f: (0, +oo) _,JR;
(b) f is 1-1, onto, and continuous on (0, +oo);
( c) f is strictly increasing if a > 1, and strictly decreasing if 0 < a < 1;
(d) loga 1 = O;
( e) Suppose a > 1. Then loga x > 0 if x > 1, and log a x < 0 if 0 < x < 1;
(f) SupposeO <a< 1. Thenlogax < 0 ifx>1, andlogax > 0 ifO < x < l;
(g) 't/x E JR, loga (ax) = x;
(h) 't/x > 0, aloga x = x.
Proof. Exercise 13. •
Theorem 5.6.24 (Properties of Logarithms, II) Let a> 0, a f-1. The
function f ( x) = loga x satisfies the following algebraic identities, 't/x, y > 0, and
rER
(a) loga(xy) = loga x + loga y;
(b) loga(x/y) = loga x - loga y;
Proof. Exercise 14. •
(c) loga (~) = -logax;
(d) loga (xr) = r loga x.