Limit as $x \to \infty$ of $\frac{\log(x)^{\log(\log(x))}}{x}$ 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 $0$?
 A: HINT:
Let $x=\exp(e^u)$. Then your limit is equal to $$\lim_{u\to\infty}\frac{(e^u)^{\log(e^u)}}{\exp(e^u)}=\lim_{u\to\infty}\frac{e^{u^2}}{e^{e^u}}=\lim_{u\to\infty}e^{u^2-e^u}=\cdots$$
A: HINT
Assume $x=e^y$ with $y \to \infty$ then
$$\frac{\log(x)^{\log(\log(x))}}{x}=\frac{\log(e^y)^{\log(\log(e^y))}}{e^y}=\frac{y^{\log y}}{e^y}=\frac{e^{\log^2 y}}{e^y}=e^{\log^2y -y}$$
A: 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 $\lim\limits_{y\rightarrow -\infty} e^y = 0$.
A: 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.
