What is $\lim\limits_{n\to\infty}\frac{n^\sqrt n}{n!}$? I am stuck at this question where I have to calculate what is big O of what,
$n!$ and $n^\sqrt n$
I tried replacing n! by it's equivalent formula but it makes everything more complicated, I can't even think about doing it by induction.
Any help would be appreciated
 A: Note that $n!\ge \left(\frac{n}{2}\right)^{n/2}$.  Hence, we have
$$\begin{align}
\frac{n^{\sqrt{n}}}{n!}&\le \frac{n^{\sqrt n}}{(n/2)^{n/2}}\\\\
&=\left(\frac{2}{n^{1-2/\sqrt{n}}}\right)^{n/2}\\\\
&\to 0\,\,\text{as}\,\,n\to \infty
\end{align}$$
A: A result I got 
$n!/ n ^ sqrt(n) $
~ $\sqrt{2*pi * n} * (n/e)^n   /  n^ \sqrt n$
= $\sqrt{2 * pi } * \sqrt n  * (n/e)^n  / n^\sqrt n $
= $\sqrt{2 * pi}  * \sqrt n * n^n / n^\sqrt n * e^n $ 
= $\sqrt{2 * pi} * n^{n+1/2} / n^\sqrt n * e^n $
= $\sqrt{2*pi}   / n^{\sqrt n-n-1/2} *e^n$
= ?
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\ln\pars{n^{\root{n}} \over n!} & = \root{n}\ln\pars{n} - \ln\pars{n!} \sim
\root{n}\ln\pars{n} - \bracks{n\ln\pars{n} - n}\quad\mbox{as}\quad n \to \infty
\end{align}
such that
$\ds{\lim_{n \to \infty}\ln\pars{n^{\root{n}} \over n!} = -\infty
\implies
\bbx{\ds{\lim_{n \to \infty}{n^{\root{n}} \over n!} = 0}}}$
