Solving l'Hôpital's Rule: Limit x→0+ (ln(x))^x

[ganondorf29]

Homework Statement

Find the limit as x->0+ of (ln(x))^x

*The answer is 1*

Homework Equations

l'Hôpital's rule

The Attempt at a Solution

lim (ln(x))^x = 0^0

I took the ln of that quantity to bring down the x

lim = x*ln(ln(x))

lim = ln(ln(x)) / (1/x)

Then I used l'Hôpital's rule

1/(x*ln(x)/(1/x^2)

= 1/(x^3*ln(x))

I got stuck here. If I plug in zero I get 0^3 and a undefined answer in the denominator. Do I have to do l'Hôpital's rule on the bottom again?

 

[Mark44]

It seems that you edited this problem while I was composing an answer. The first problem I saw was lim (-ln(x))^x, as x approached 0+.
A little while later, the minus sign in front of ln(x) was gone. Here's my work for the original problem. For the edited problem, no limit exists.

$$Let y = (-ln x)^x$$
$$ln y = x ln(-ln(x)) = \frac{ln(-ln(x))}{\frac{1}{x}}$$

$$lim_{x \rightarrow 0^+} ln y = lim_{x \rightarrow 0^+} \frac{ln(-ln(x))}{\frac{1}{x}} [\frac{\infty}{\infty}]$$

$$= lim_{x \rightarrow 0^+} \frac{\frac{1}{-ln(x)}\frac{-1}{x}}{\frac{-1}{x^2}}$$
$$= lim_{x \rightarrow 0^+} \frac{-x}{ln(x)} = 0$$
Since $$lim_{x \rightarrow 0^+} ln y = 0$$
Then $$ln(lim_{x \rightarrow 0^+} y) = 0$$
Or $$lim_{x \rightarrow 0^+} (-ln(x))^x = 1$$