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Evaluate the following limit.

$$ \lim_{x\to \infty} (\ln\ x)^{\frac{1}{x}} $$

What i have tried:

$$ \ln\left[\lim_{x\to \infty} (\ln\ x)^{\frac{1}{x}}\right] $$

$$ \lim_{x\to \infty} \ln(\ln x)^{\frac{1}{x}} $$

$$ \lim_{x\to \infty} \frac{\ln(\ln x)}{x} $$

So as $ x$ approaches infinity, the limit goes to 0. But the answer in the book is 1.

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    $\begingroup$ The limit of the logarithm is $0$. So what's the original limit? $\endgroup$ – David Mitra Feb 6 '14 at 21:53
  • $\begingroup$ Could you use L'Hospital? $\endgroup$ – user122283 Feb 6 '14 at 21:58
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    $\begingroup$ Could you try rewriting $(\ln x)^{1/x}$ as $\exp(\ln(\ln x)/x)$? $\endgroup$ – TooTone Feb 6 '14 at 22:06
  • $\begingroup$ If any of the answeres below were useful to you, then you should upvote all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this. $\endgroup$ – gebruiker Mar 21 '16 at 18:48
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You took the natural log $\ln$ of the limit to evaluate it easier, but you forgot to undo the natural log. It is just like how if you were to add $1$ to the limit to make it easier to calculate, you would have to subtract off $1$ in the end.

In this case, to "undo" a natural log, you take $e$ to the power of something. So after you took the natural log you calculated the limit to be $0$; then $$ e^0 = 1 $$ is your answer.

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I am still learning about limits, but I think the following approach ought to work.

For any $p,q$, $\;p^q = \exp(\ln(p)q)$. Hence $$\ln(x)^{1/x} = \exp\left(\frac{\ln(\ln x)}{x}\right)$$ And taking limits $$\begin{align}\lim_{x\to\infty} \ln(x)^{1/x} &= \lim_{x\to\infty} \left( \exp\left(\frac{\ln(\ln x)}{x}\right) \right)\\ &= \exp\left(\lim_{x\to\infty}\left(\frac{\ln(\ln x)}{x}\right) \right) \end{align}$$

and hopefully you can see the inner limit evaluates to 0.

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