How to solve this limit problem?-$\lim_{n\to \infty}\ \left(\frac{\ n!}{(mn)^n}\right)^{\frac{1}{n}}$ I need to find the value of-
$$\lim_{n\to \infty}\ \left(\frac{\ n!}{(mn)^n}\right)^{\frac{1}{n}}$$
where $m {\in} R$
I don't know how to even start. Would someone explain it step by step, also which type of indeterminate form is this?
Is there a simpler to solve this ? i.e. without any high mathematics theorem etc.?
 A: Use Stirling's approximation for $n!$
$n!\approx \sqrt{2\pi n}( \frac ne)^n$
$$\lim_{n\to \infty} \left(\frac{\sqrt{2\pi n}}{(em)^n}\right)^{\frac 1n}$$
$$\frac {1}{em} \lim_{n\to \infty} (2\pi n)^{\frac 1{2n}}$$
Let the above limit (excluding constant) be $L$
$$\log L= \lim_{n\to \infty}\frac{\log (2\pi n)}{2n}$$
L'Hopital's rule here
A: You don't need Stirling.
$ \left(\dfrac{\ n!}{(mn)^n}\right)^{\frac{1}{n}}
=\dfrac{(n!)^{1/n}}{mn}
$
so all you need is that
$\dfrac{(n!)^{1/n}}{n}
\to \dfrac{1}{e}
$
to get that the limit is
$\dfrac1{em}
$.
$\dfrac{(n!)^{1/n}}{n}
\to \dfrac{1}{e}
$
follows from 
$\left(\dfrac{n}{e}\right)^n
< n!
<\left(\dfrac{n}{e}\right)^{n+1}
$.
These, in turn,
can be proved by induction from
$\left(1+\dfrac1{n}\right)^n
< e
<\left(1+\dfrac1{n}\right)^{n+1}
$.
A: let $a_n = \frac {n!}{(mn)^n}$.
by the root-frac analogy we know that for every $a_n$ if the limit of the series $b_n = \frac{a_{n+1}}{a_n}$ exist it is equal to the limit of the series $c_n = \sqrt[n]{a_n}$. so lets calculate the limit of the series $b_n = \frac{a_{n+1}}{a_n}$
$b_n = \frac{a_{n+1}}{a_n} =  \frac {(n+1)!}{(m((n+1)))^{(n+1)}} : \frac {n!}{(mn)^n} = m^{-1}(\frac{n}{n+1})^n$
we know that the limit of $(\frac{n}{n+1})^n$ (n goes to infinity) is $e^{-1}$ then the limit of $b_n$ (since m is a constant different than 0) is $m^{-1}e^{-1} = \frac{1}{me}$
final result: $\frac{1}{me}$
A: $$\lim_{n\to\infty}\left(\frac{n!}{(mn)^n}\right)^{\frac{1}{n}}=\lim_{n\to\infty}\exp\left[\ln\left(\left(\frac{n!}{(mn)^n}\right)^{\frac{1}{n}}\right)\right]=$$
$$\lim_{n\to\infty}\exp\left[\frac{\ln\left(\frac{n!}{(mn)^n}\right)}{n}\right]=\exp\left[\lim_{n\to\infty}\frac{\ln\left(\frac{n!}{(mn)^n}\right)}{n}\right]=$$

Apply l'Hôpital's rule:

$$\exp\left[\lim_{n\to\infty}\frac{\psi^{(0)}(1+n)-\ln(mn)-1}{1}\right]=$$
$$\exp\left[\lim_{n\to\infty}\left(\psi^{(0)}(1+n)-\ln(mn)-1\right)\right]=$$
$$\exp\left[\left(0-\ln(m)-1\right)\right]=\exp\left[-\ln(m)-1\right]=\frac{1}{me}$$


*

*To find this limit $\lim_{n\to\infty}\left(\psi^{(0)}(1+n)-\ln(mn)-1\right)$, notice:


$$\lim_{n\to\infty}\left(\psi^{(0)}(1+n)-\ln(mn)-1\right)=$$
$$\lim_{n\to\infty}\left(\psi^{(0)}(1+n)-\ln(mn)\right)-1=$$
$$\lim_{n\to\infty}\ln\left(\frac{\exp\left[\psi^{(0)}(1+n)\right]}{mn}\right)-1=\ln\left(\frac{1}{m}\right)-1$$
