1549901369-Elements_of_Real_Analysis__Denlinger_

5.6 *Exponentials, Powers, and Logarithms 285

Hence sup{ xt : x < xo} > Xb - c. By the forcing principle, this implies

sup{xt : x < xo} ~ x6. Therefore, sup{xt : x < xo} = x6.

(2) Next, we must prove that inf{xt: x > x 0 } = x6. (Exercise 8)

Summarizing: in Case 1, lim f(x) = f(x 0 ) = lim f(x). Therefore, the

X->Xo X->xci

power function f ( x) = xt is continuous at every x 0 ~ 1.

Case 2 (0 < xo < 1):

Then -^1 > l. Thus, by Case 1, lim xt = -( 1 )t. Let {xn} be any mono-

xo X-> xlo Xo

tone decreasing sequence of real numbers converging to x 0. Then { Xln } is

1.

a monotone increasing sequence converging to -. As proved m Case 1, the

power function f ( x) = xt is continuous at ~ x:o ( 2_) t ___, ( ~) t That
xo Xn xo
1 1
is, t ---> t"· Thus, by the algebra of limits, x~ ---> x6. Therefore, the power
Xn Xo
function f ( x) = xt is continuous at every 0 :::; xo < 1.

(d) Next we prove that X-> lim xt = +oo. By the density of the rational
m
numbers in JR, :J m,n E N 3 m > 1 and 0 < < t. Let M > 0. Since
n
lim xm = +oo (by Theorem 4.4.18) :J xo > 1 3
X->

x 0 >Mn.

Then, since the power function g(x) = x^1 1n is strictly increasing (Exercise
5.5.15),
x~/n > Mn/n = M.

Since the exponential function h(u) = xg is strictly increasing (Theorem 5.6.6),

xo t > Xo m/n > M.

Finally, since the power function f ( x) = xt is strictly increasing (proved in ( c)
above),
x > xo ==> xt > Xb ==> xt > M.

Therefore, lim xt = +oo.
X->
(e) Using Theorem 4.4.21 (a),
1 1 1
lim xt = lim - - = lim - = lim --= 0. •
x->O+ l/x->= (1/x)t u->= ut u-> f(u)