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 $x \rightarrow 0^+$.