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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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Coming from the right suggests that $x$ stays positive, so the $\ln$ will be well-defined. The power suggests we use logarithmic differentiation, which says something to the effect of $\lim f(x) = \exp(\lim \ln f(x))$. Do you see why that's useful? – Eugene Shvarts Nov 14 '12 at 14:14
That is what I did. But I am having trouble calculating the limit of ln(ln(1/x)) – user43418 Nov 14 '12 at 14:15
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. – Eugene Shvarts Nov 14 '12 at 14:18

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 $0$.

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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 +1 – Adi Dani Mar 4 '13 at 18:01

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