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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

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