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I want to compute $\lim\limits_{x \to \infty}\frac{\log(x)^{\log(\log(x))}}{x}$

By graphing it, clearly $x$ grows larger than $\log(x)^{\log(\log(x))}$, so the limit will go to $0$.

I tried iterating L'Hopital's rule, but after three derivations, the sequence of limits gets successively more complicated.

How can you prove that the limit is indeed $0$?

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  • $\begingroup$ Maybe a change of variable like $t = \log(x)$ would make it a bit easier? $\endgroup$ – xbh Aug 29 '18 at 6:39
  • $\begingroup$ $$\log(f(e^z))=\log^2 z-z\to-\infty$$ $\endgroup$ – Yves Daoust Aug 29 '18 at 7:28
  • $\begingroup$ I think a parentheses should be placed outside the "base" $\log$, otherwise to find the limit is relatively easy. $\endgroup$ – dmtri Aug 30 '18 at 5:54
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HINT:

Let $x=\exp(e^u)$. Then your limit is equal to $$\lim_{u\to\infty}\frac{(e^u)^{\log(e^u)}}{\exp(e^u)}=\lim_{u\to\infty}\frac{e^{u^2}}{e^{e^u}}=\lim_{u\to\infty}e^{u^2-e^u}=\cdots$$

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HINT

Assume $x=e^y$ with $y \to \infty$ then

$$\frac{\log(x)^{\log(\log(x))}}{x}=\frac{\log(e^y)^{\log(\log(e^y))}}{e^y}=\frac{y^{\log y}}{e^y}=\frac{e^{\log^2 y}}{e^y}=e^{\log^2y -y}$$

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Take logarithm!

Then you are interested in the limit $(\log \log x)^2-\log x$, which is clearly $-\infty$. So the answer equals to the limit $\lim\limits_{y\rightarrow -\infty} e^y = 0$.

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We can implement L'Hopital's rule:

$A= \lim\limits_{x \to \infty}\frac{log(x)^{log(log(x))}}{x}$, according to $log$ rule: $log A ^ B=BlogA$ we have:

$A = \lim\limits_{x \to \infty} \frac {log(log(x))log(x)}{x}$ which is rule type $\frac{\infty}{\infty}$ so, we can implement L'Hopital's rule

$A=\lim\limits_{x \to \infty}\frac{\frac{1}{log(x)}\frac{1}{x}log(x)+log(log(x))\frac{1}{x}}{1}$, then,

$A=\lim\limits_{x \to \infty}(\frac{1}{x}+\frac{log(log(x))}{x})$, and, by $\lim\limits_{x \to \infty}\frac{1}{x}=0$, we have:

$A=\lim\limits_{x \to \infty}\frac{log(log(x))}{x}$, which is rule type $\frac{\infty}{\infty}$ so, we can implement L'Hopital's rule again:

$A=\lim\limits_{x \to \infty}\frac{\frac{1}{log(x)}\frac{1}{x}}{1}=\lim\limits_{x \to \infty}\frac{1}{x log(x)}$, which is limit rule $\frac{1}{\infty}=0$, so, finaly we have:

$A=0$

Thanks, Vladica.

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  • $\begingroup$ I think a different limit is needed....you may see answers above $\endgroup$ – dmtri Aug 29 '18 at 15:47
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    $\begingroup$ I don't think this solution is valid, the numerator is in the form $\log(a)^b$, but the log rule only applies for equations of the form $\log(a^b)$ $\endgroup$ – user477805 Aug 30 '18 at 1:57
  • $\begingroup$ in that case, should be $(log(a))^b$ or $log^b(a)$ $\endgroup$ – vladadj31 Aug 30 '18 at 7:07

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