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So I'm stuck with this:

$$\lim_{x \to \infty} (\log x)^{1/x}$$

I think it goes to $1$.

Actually the limit I'm trying to solve is a little bit more complicated, but it boils down to this.

(It's been some time since I've worked with this kind of limit problems, but I think I should know where to start, but nope ...)

Hints preferred first.

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  • $\begingroup$ To resolve exponential indeterminate cases of the limit us that $f^g=e^{g\log(f)}$ to reduce to the study of the limit of $g\log(f)$. In your case of $\frac{\log(\log(x))}{x}$ $\endgroup$ – Pp.. Feb 8 '15 at 12:38
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Hint

If $$A=(\log x)^{1/x}$$ $$\log(A)=\frac{\log\big(\log(x)\big)}{x}=\frac uv$$ and use L'Hopital : $u'=\frac{1}{x \log (x)}$, $v'=1$. Then the limit of $\log(A)$ is the same as the limit of $\frac{1}{x \log (x)}$ which is $0$ when $x$ goes to $\infty$. So the limit of $A$ is $1$.

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$\bf hint:$

let $y = (\ln x)^{1/x}.$ then $\ln y = \dfrac{\ln( \ln x)}{x}.$ take the limit of this quotient.

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Hint: $f(x)=\log(\log x)/x\to 0$ as $x\to \infty$.

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Write $f(x)^{g(x)}$ as $\exp\left(\frac{\log(f(x))}{g(x)}\right)$:

$$\lim_{x \to +\infty}(\log x)^{1/x} = \lim_{x \to +\infty}\exp\left(\frac{\log(\log x)}{x}\right) = \exp\left(\lim_{x \to +\infty}\frac{\log(\log x)}x\right) = 1$$

because $e^x$ is continuous and $$\frac{\log(\log x)}x < \frac{\log x}x \xrightarrow[\quad\infty]{} 0.$$

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