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Compute the limit of the given function

Evaluate the limit.

    \[ \lim_{x \to 0^+} \left( \log \left( \frac{1}{x} \right) \right)^x. \]


First, we note that

    \[ \log \left( \frac{1}{x}\right) = \log x^{-1} = - \log x. \]

Then we use the continuity of the exponential,

    \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*}

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