1549901369-Elements_of_Real_Analysis__Denlinger_

6.6 *L'Hopital's Rule 353

We can use L'Hopital's rule a second time, this time, because we have the
indeterminate form 0/0. We get

lim x ( - sin x) + cos x = 0 + 1 = 1.

x-->0 COS X 1

D

OTHER INDETERMINATE FORMS

Sometimes other indeterminate forms such as oo - oo, 0 · oo, o^0 , etc. can
be evaluated using L'Hopital's rule, as illustrated in the following example.

Examples 6.6.8 Find each of t he following limits.

(a) hm. ( 1 -. - - -1)

x-->O smx x

(b) lim x lnx

x-->O+

( c) lim xx [where we define xx = ex ln x ].

x-->O+

Solution: (a) This is an example of the form oo - oo. Notice that

1 Im. ( --^1 - -1) = 1 Im. x - sin x.

x-->O sin x x x-->O x sin x

which is now of the form 0/0. Using L 'Hopital's rule (twice) we have

1 - cosx sinx 0

lim. = lim - --- 0

x-->O XCOSX + S!IlX x-->O -xsinx + 2COSX - 0 + 2 -.

(b) This is an example of the form 0 · - oo.

lnx l/x

li m xlnx = lim - = lim --- 2

x-->O+ x-->O+ 1 / X x-->O+ -1 / X

2

= lim - x = lim (-x) = 0.

x-->O+ X x-->O+

( c) This is an example of the form o^0. We shall use the logarithm function to
convert the expression xx to a nother form. Let f ( x) = xx. Then ln f ( x) = x In x,
and lim lnf(x) = lim xlnx = 0, by P art (b).
x-->O+ x-->O+
But ex is continuous on JR; thus by Theorem 5.1.14 (b),
lim e^1 nf(x) = elimlnf(x); that is,
x-->O+
lim f(x) = e^0 = l ; i.e., lim xx= 1. D
x-->O+ x-->O+