1549901369-Elements_of_Real_Analysis__Denlinger_

348 Chapter 6 11 Differentiable Functions

(b) As x-> 1, lnx-> 0 and x -1-> 0. Thus, by Theorem 6.6.4,

lnx 1 /x

lim --= lim - = 1.

x-+l X - 1 x-+l 1

We could have evaluated this limit without L'Hopital's rule. Writing the

numerator as lnx - lnl, we see that this limit is just the derivative of lnx at

x = 1.

(c) As x-> 7r/2, 1-sinx-> 0 and cos^2 x-> 0. Thus, by Theorem 6.6.4,

1

. 1-sinx
1
. -cosx
1
. 1 1
im = im = im --= -.
x-+rr /2 cos^2 x x-->rr /2 2 cos x ( - sin x) x-+rr /2 2 sin x 2

As in (b), L'Hopital's rule is not really needed to find this limit. We can

evaluate it using a trigonometric identity, and the algebra of limits:

1

. 1 - sin x
1
. 1 - sin x
1

.

lm = lm 2 = lm

x-->rr /2 cos^2 x x-->rr /2 1 - sin x x-->rr /2 1 + sin x

1 1

2

( d) As x -> 7r /2, 1 +sin x ft 0. Thus, L 'Hopital's rule cannot be used here.

We can, however , easily evaluate this limit. As x -> 7r/2, 1 + sinx -> 2 and

cos^2 x -> 0, while cos^2 x > 0. Thus,

1

. 1 + sinx
im = +oo.
x-+rr ;2 cos^2 x

Notice that if we mistakenly apply L'Hopital's rule to evaluate this limit, we
cosx -1 -1
would get the incorrect answer, lim (. ) = lim -.- -
x-->rr /2 2 cos x - sm x x -->rr /2 2 sm x 2
0

Theorem 6.6.6 (L'Hopital's Rule II, for f /oo) Suppose f , g : I-> JR,

where I is an open interval with "endpoint" a, and where

(a) a may be finite, +oo or -oo;

(b) f and g are differentiable on I;

(c) \Ix E J, g(x)g'(x) -j. O;

(d) lim g(x) = +oo or -oo;

X-->Q

. f'(x).
(e) Xhm -->a ~( g X ) = L (finite, +oo or -oo).
. f(x). f'(x)

Then X-->Q hm -(-) g X = X-->Q hm -(-) g (^1) X = L.