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I am having a problem with the calculation of the following limit.

I need to find: $\lim_{x \to 0^{+}} (\ln \frac{1}{x})^x$

Thank you in advance

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    Naively, as $x\to 0$, for positive $x$, $1/x \to \infty$, and as $u\to\infty$, $\ln u \to \infty$ as well. Taking another $\ln$ only compounds the problem. But, that's not the only term in the limit; having infinities involved generally means L'Hôpital is afoot.2012-11-14

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Take $f(x)=\ln^x(1/x)$ so as @Eugene suggested take $\ln$ of both sides, so $\ln(f(x))=x\ln(\ln(1/x))$ or $\ln(f(x))=\frac{\ln(\ln(1/x))}{1/x}$ Now take $1/x=t$ so when $x$ tends to $0^+$; $t$ tends to $+\infty$. By L'Hospital's Rule you have $\ln(f(x))\to 0$ so $f(x)\to 1$.

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    Nice observation Babak +12013-03-04
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It may be easier to look at large numbers. So let $w=1/x$. We study $(\ln w)^{1/w}$. Take the logarithm of this. We get $\dfrac{\ln(\ln w)}{w}$.

By L'Hospital's Rule, or otherwise, this approaches $0$ as $w\to\infty$. So our limit is $1$.

For the L'Hospital's Rule argument, differentiating top and bottom yields $\dfrac{1}{w\ln w}$, which clearly has limit $0$.