282 Chapter 5 • Continuous FunctionsLemma 5.6.9 Let a ;::: 1 and x E R If {tn} is any monotone decreasing
sequence of rational numbers converging to x, then lim atn =ax.
n-+OO
Proof. Let a,x, and {tn} satisfy the hypotheses. Define the sequence {rn}
by rn = 2x - tn. Then, {rn} is a monotone increasing sequence of rational
numbers and Tn --) x, So, by Definition 5.6.5, arn --) ax. Using the properties
of ax obtained in Theorem 5.6.8,
a2x
lim atn = lim a^2 x-r,. = lim -
n-+OO n---tOO n-+OO arn
a2x a2x
lim arn ax
n-+oo
Theorem 5.6. 10 Let a> 0 and x ER If {xn} is any sequence of real numbers
converging to x, then lim axn = ax.
n-+oo
Proof. Suppose a, x, and {xn} satisfy the hypotheses.
Case 1 (a ;::: 1): We know that :3 strictly increasing sequence { r n} of rational
numbers and :3 strictly decreasing sequence { sn} of rational numbers such that
rn --) x and Sn --) x. Then, by Definition 5.6.5 and Lemma 5.6.9,
n-+oo n-+ooAlso, in what follows, recall that the exponential function f(x) =ax is strictly
increasing.
Let c: > 0. Then, since rn--) x and Sn--) x , :3 n 1 EN 3
Note that rn, < x < sn,. Thus, since Xn--) x, :3 no EN 3
Since f(x) =ax is strictly increasing, this means,
Therefore, lim axn = ax.
n-+oo
Case 2 (0 <a< 1) : Then a-^1 > 1 and by Case 1,
1 1- ---- - ---- ax
lim (a-l)xn - (a-1 )x - ·
n-+oo
lim axn = lim l
n--+oo n--+oo (a-1 )Xn •