1549901369-Elements_of_Real_Analysis__Denlinger_

6.6 *L'Hopital's Rule 347

a change of variables and introduce two new functions. We define functions F

and G on (0, ~) by

F(u) = { f (~) ~f 0 < u < ~ } and G(u) = { g (~) ~f 0 < u < ~ }.

0 du=O 0 du=O

By the chain rule, F and Gare differentiable on (0, ~).By Theorem 4.4. 19 ,

u-+O lim F(u) = u-+O lim f (~) U = x-++oo lim f(x) = 0,

and similarly, lim G(u) = 0.

u-+O

(

1) -1 -!' (1-) -g' (1·)

By the chain rule, F'(u) = f' - · - = u and G'(u) = u.

u u^2 u^2 u^2

Thus, G'(u) -:/:- 0 on (o, ~). Therefore, F and G satisfy all the hypotheses of

Case 1, 2, or 3 for the interval (0, ~).Hence,

f (.!.) F(u) F'(u)

lim __ u_ = lim -- lim --

u-+O+ g (~) u-+O+ G(u) u-+O+ G'(u)

. -!' (~) /u2. !' (~)

= u-+O+ hm -g' (u 1 ) /u 2 = u-+O+ lim -(g' (^1) u ).
That is (see Theorem 4.4.19 and Exercise 4.4-B.14),
lim f(x) = lim f'(x).
x-++oo g( X) x-++oo g' ( X)
Cases 13, 14, and 15: a= -oo, and L = a (finite) real number + oo, or - oo.
(Exercise 5) •
Examples 6.6.5 Calculate each of the following limits. Before using L'Hopital's
rule, be sure that the hypotheses are met.
sinx. lnx
(a) lim - 2 (b) hm --
x-+O+ X X-+l X - 1
1

. 1 - sinx
(c) Im
x-+11" ;2 cos^2 x

(d) lim 1 + sinx

x-+11" /2 cos^2 x

Solution: (a) As x-+ o+, sinx-+ 0 and x^2 -+ 0. Thus, by Theorem 6.6.4,

1

. sin x
1
. cos x
Im --= Im --= +oo
x-+O+ x^2 x-+O+ 2x

(As x-+ o+, cosx-+ 1, 2x-+ 0, and x > 0.)