Vtyjkyuk

I want to compute $limlimits_x to inftyfraclog(x)^log(log(x))x$

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 $0$?

HINT:

Let $x=exp(e^u)$. Then your limit is equal to $$lim_utoinftyfrac(e^u)^log(e^u)exp(e^u)=lim_utoinftyfrace^u^2e^e^u=lim_utoinftye^u^2-e^u=cdots$$

HINT

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

$$fraclog(x)^log(log(x))x=fraclog(e^y)^log(log(e^y))e^y=fracy^log ye^y=frace^log^2 ye^y=e^log^2y -y$$

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 $limlimits_yrightarrow -infty e^y = 0$.

We can implement L'Hopital's rule:

$A= limlimits_x to inftyfraclog(x)^log(log(x))x$, according to $log$ rule: $log A ^ B=BlogA$ we have:

$A = limlimits_x to infty frac log(log(x))log(x)x$ which is rule type $fracinftyinfty$ so, we can implement L'Hopital's rule

$A=limlimits_x to inftyfracfrac1log(x)frac1xlog(x)+log(log(x))frac1x1$, then,

$A=limlimits_x to infty(frac1x+fraclog(log(x))x)$, and, by $limlimits_x to inftyfrac1x=0$, we have:

$A=limlimits_x to inftyfraclog(log(x))x$, which is rule type $fracinftyinfty$ so, we can implement L'Hopital's rule again:

$A=limlimits_x to inftyfracfrac1log(x)frac1x1=limlimits_x to inftyfrac1x log(x)$, which is limit rule $frac1infty=0$, so, finaly we have:

$A=0$

Thanks, Vladica.