1549901369-Elements_of_Real_Analysis__Denlinger_

284 Chapter 5 • Continuous Functions

Since limits preserve inequalities,

xt :=::; qi :=::; q~ :=::; yt.

By Exercise 5.5.15, qi < q~. Therefore

xt < yt.

(c) We next show that the power function f(x) = xt is continuous at every

positive number x 0. Let x 0 > 0. Since f is strictly increasing on (0, +oo ), we

have by Theorem 5.2.17,

lim -f(x) = sup{xt: x < xo}:::; f(xo):::; inf{xt: x > xo} = lim + f(x).

X---tXo X---+Xo

We wish to prove that lim f(x) = f(xo) = lim f(x); that is,

X---+Xo - X---tXa +

sup{xt: x < xo} = x6 = inf{xt: x > xo}.

Case 1 (xo 2: 1):

(1) First, we prove that sup{xt : x < x 0 } = x&. Let {rn} be a monotone

increasing sequence of rational numbers such that rn -t t. By Definition 5.6.5,

Let c: > 0. Then, since {x~n} is monotone increasing, 3 n 0 EN 3

n 2: no ::::} Xb - c: < x~n :=::; x&

(16)

Fix an integer n 2: no. By Exercise 5.5.15 (c), the power function gn(x) = xrn

is continuous at x 0. Thus, 3 o > 0 3

[See (16).]
So,

Pick any x E (xo - o, xo). Since the function f(x) = xt is strictly increasing,
and rn < t,

(18)

Putting (17) and (18) together,

Xo -o < x < Xo ::::} x& -c: < xt = f ( x).