Given limit, I am confused with L'hopital's rules here due to existing $\ln$
$$\lim_{x \to 0^+}\Big(\ln\frac{1}{x}\Big)^x$$
how to find the limit of $\lim_{x \to 0^+}\Big(\ln\frac{1}{x}\Big)^x$?
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limits
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3Please include additional context. Perhaps explain what you know about l'Hopital's rule and why this doesn't fit the rule. – 2017-02-23
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1Try this $\ln (1/x) = e^{\ln (\ln (1/x))}$. Now use L'Hopital's rule in the exponent. – 2017-02-23
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0note that $$\ln\left(\frac{1}{x}\right)=\ln(1)-\ln(x)=-\ln(x)$$ – 2017-02-23
1 Answers
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Hint: write your limit in the form $$e^{\frac{\ln(-\ln(x))}{\frac{1}{x}}}$$ and use L'Hospital