Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I formally show that the following limit is $0$? $$\lim_{x\rightarrow 0^+} x^{-\ln x}$$ (Without using l'Hopital's rule.)

I can write it as $$\lim_{x\rightarrow 0^+} x^{-\ln x} = \lim_{x\rightarrow 0^+} e^{-\ln x \ln x}.$$ I would somehow need to argue that $ - \ln x \ln x \rightarrow -\infty$ as $x \rightarrow 0^+$.

share|cite|improve this question
As $x \downarrow 0$, $\ln x \to -\infty$, so $(\ln x)^2 \to \infty$, so $-(\ln x)^2 \to -\infty$. – copper.hat Nov 6 '12 at 18:42
up vote 2 down vote accepted

For the last part of your argument, informally, you can note that $\ln x \to -\infty$ as $x\to 0$. Then you have

$$-(-\infty \cdot -\infty) = - (\infty) = -\infty$$

share|cite|improve this answer
Or even formally if phrased in terms of limit forms. Or better, if you use the extended real numbers, and the fact the (continuous extension of) the logarithm is continuous on $[0,+\infty]$. – Hurkyl Nov 6 '12 at 19:27
Ah, nice thinking. That was actually what I was thinking once I thought about this for a few minutes. My original answer was rushed as I was just about to leave for class. – Joe Nov 6 '12 at 22:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.