# Show $\lim_{x\rightarrow 0^+} x^{-\ln x} = 0$ without L'Hopital

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 $x \rightarrow 0^+$.

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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

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