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

Question

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 $x \rightarrow 0^+$.

Answer 1

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

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