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Section 8.2 L’Hôpital’s Rule 449<br />
We apply l’Hôpital’s Rule to the previous expression.<br />
lim ln f x lim<br />
x→ x→<br />
lim<br />
x→<br />
ln ( 1 1 x ) <br />
<br />
1 x <br />
(<br />
1<br />
x 2)<br />
1<br />
<br />
1 1 x <br />
<br />
x<br />
12 <br />
0 0 <br />
Differentiate numerator<br />
and denominator.<br />
Therefore,<br />
1<br />
lim<br />
1<br />
x→<br />
1 1 x <br />
x<br />
lim<br />
x→ ( 1 1 x ) lim f x lim e ln f x e 1 e.<br />
x→ x→<br />
Now try Exercise 21.<br />
EXAMPLE 9 Working with Indeterminate Form 0 0<br />
Determine whether lim x→0<br />
x x<br />
exists and find its value if it does.<br />
[–3, 3] by [–1, 3]<br />
Figure 8.11 The graph of y x x .<br />
(Example 9)<br />
SOLUTION<br />
Investigate Graphically Figure 8.11 suggests that the limit exists and has a value<br />
near 1.<br />
Solve Analytically The limit leads to the indeterminate form 0 0 . To convert the<br />
problem to one involving 00, we let f x x x and take the logarithm of both sides.<br />
ln f x x ln x l n x<br />
<br />
1x<br />
Applying l’Hôpital’s Rule to ln x1x we obtain<br />
lim ln f x lim l n x<br />
<br />
x→0 x→0 1x<br />
lim 1x<br />
<br />
x→0 <br />
1x 2<br />
lim x 0.<br />
x→0 <br />
Therefore,<br />
<br />
<br />
<br />
Differentiate.<br />
lim x x lim f x lim<br />
x→0 x→0 x→0 eln f x e 0 1.<br />
Now try Exercise 23.<br />
EXAMPLE 10 Working with Indeterminate Form 0<br />
Find lim<br />
x→<br />
x 1x .<br />
SOLUTION<br />
Let f x x 1x . Then<br />
ln f x ln x<br />
.<br />
x<br />
continued