Calculate the limit $\lim _{n \rightarrow \infty} \frac{[\ln (n)]^{2}}{n^{\frac{1}{\ln (\ln (n))}}}$.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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 $0$.