Can be done without L’Hopital: we need a substitution y = -log(x), and then expand the fraction log(y)/e^{y} into log(y)/y * y/e^y, so that Theorem 7.11 can be applied
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![\[ \lim_{x \to 0^+} \left( \log \left( \frac{1}{x} \right) \right)^x. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-2c018e35843fc3ced0790ad1b376efe1_l3.png "Rendered by QuickLaTeX.com")
![\[ \log \left( \frac{1}{x}\right) = \log x^{-1} = - \log x. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-6c0de26527f1dcd6993956b16e44c953_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} \lim_{x \to 0^+} \left( \log \left( \frac{1}{x} \right) \right)^x &= \lim_{x \to 0^+} (-\log x)^x \\[9pt] &= \lim_{x \to 0^+} e^{- x \log (\log x)} \\[9pt] &= \exp \left( \lim_{x \to 0^+} (-x \log (\log x)) \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{ \log (\log x)}{\frac{1}{x}} \right)\\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{ -\frac{1}{x \log x}}{-\frac{1}{x^2}} \right) & (\text{L'Hopital}) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{x}{\log x} \right) \\[9pt] &= \exp \left( \lim_{x \to 0^+} \frac{1}{\frac{1}{x}} \right)&(\text{L'Hopital}) \\[9pt] &= e^0 = 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-228fa614f71767336ed91f839b92b116_l3.png "Rendered by QuickLaTeX.com")
Can be done without L’Hopital: we need a substitution y = -log(x), and then expand the fraction log(y)/e^{y} into log(y)/y * y/e^y, so that Theorem 7.11 can be applied