Calculate the limit $\lim _{n \rightarrow \infty} \frac{[\ln (n)]^{2}}{n^{\frac{1}{\ln (\ln (n))}}}$. I would like to know how to calculate the limit
$$
\lim _{n \to \infty}
{\ln^{2}\left(n\right) \over n^{ 1/\ln\left(\,{\ln\left(\,{n}\,\right)}\,\right)}}
$$
I have tried to change its form using
$\exp\left(\,{\ln\left(\,{x}\,\right)}\,\right) = x$ and changing
$X = \ln(\,{x}\,)$ but it came down to computing the limit of
$$
\lim _{X \to \infty}
\left[X^{2}\mathrm{e}^{-X/\ln\left(\,{X}\,\right)}\right]
$$
Any suggestions ?. Thanks.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\lim_{n \to \infty}{\ln\pars{n} \over n^{1/\ln\pars{\ln\pars{n}}}} &
\,\,\,\stackrel{n\ \mapsto\ {\large\expo{n}}}{=}\,\,\,
\lim_{n \to \infty}{n \over \expo{n/\ln\pars{n}}}
\,\,\,\stackrel{n\ \mapsto\ {\large\expo{n}}}{=}\,\,\,
\lim_{n \to \infty}{\expo{n} \over \exp\pars{\expo{n}/n}}
\\[5mm] = &\
\lim_{n \to \infty}\exp\pars{n - {\expo{n} \over n}} =
\bbx{\large 0} \\ 
\end{align}
A: Lat
$$x=\log(\log(n)) \implies n=e^{e^x}$$ which makes the expression to be
$$A=\frac{[\ln (n)]^{2}}{n^{\frac{1}{\ln (\ln (n))}}}=e^{2 x-\frac{e^x}{x}}$$ Now, ${2 x-\frac{e^x}{x}}<0$ as soon as
$$x > -2 W\left(-\frac{1}{2 \sqrt{2}}\right)  \approx 1.5$$ and ${2 x-\frac{e^x}{x}}\to -\infty$ and then the limit of $0$.
A: By $x=\ln n \to \infty$ we have
$$\frac{[\ln (n)]^{2}}{n^{\frac{1}{\ln (\ln (n))}}} = \frac{x^{2}}{e^{\frac{x}{\ln x}}}= \frac{\left(\frac{x}{\ln x}\right)^3}{e^{\frac{x}{\ln x}}}\frac{(\ln x)^3}{x}\to 0\cdot 0=0$$
indeed by $y=\frac{x}{\ln x}\to \infty$ eventually $e^y\ge y^4$ and then
$$\frac{\left(\frac{x}{\ln x}\right)^3}{e^{\frac{x}{\ln x}}}=\frac{y^3}{e^y} \le \frac{y^3}{y^4}=\frac1y \to 0$$
and by $\ln x=z \to \infty$
$$\frac{(\ln x)^3}{x}=\frac{z^3}{e^z}\to 0$$
A: Let's look at the denominator:
$$n^{\frac{1}{\ln (\ln  n )}}=\mathrm e^{\frac{\ln n}{\ln (\ln  n)}}$$
so we can rewrite the fraction as
$$\frac{n^2 }{n^{\frac{1}{\ln (\ln (n))}}}=\mathrm e^{2\ln(\ln n)-\tfrac{\ln n}{\ln(\ln n)}}. $$
Now consider the exponent: as $\ln^2 u=_{\infty}o(u)$, we have $2\ln^2u-u\sim_\infty -u$, whence by the substitution $u=\ln n$,
$$-\frac{2(\ln(\ln n))^2-\ln n}{\ln(\ln n)}\sim_\infty-\frac{\ln n}{\ln(\ln n)}\to 0.$$
