Do you remember summer '09?

We have $\text{L}=\displaystyle \lim_{n \to \infty}\left(\dfrac{n!}{(mn)^n}\right)^{1/n}=\displaystyle \lim_{n \to \infty}\dfrac{1}{m}\left(\dfrac{1}{n}\dfrac{2}{n}...\dfrac{n}{n}\right)^{1/n}$

$\ln(L)=\displaystyle \lim_{n \to \infty}\left[\ln(\dfrac{1}{m})+\dfrac{1}{n}\left(\ln(\dfrac{1}{n})+\ln(\dfrac{2}{n})+\ln(\dfrac{3}{n})+....+\ln(\dfrac{n}{n})\right)\right]=-\ln(m)+\displaystyle \lim_{n \to \infty}\dfrac{1}{n}\displaystyle\sum_{r=1}^n \ln\left(\dfrac{r}{n}\right)$

$=-\ln(m)+\displaystyle\int_0^1 \ln(x)\ dx=-\ln(m)+\left[x(\ln(x)-1)\right]_0^1=-\ln(m)-1=\ln(\frac{1}{em})$

By stirling approximation the above expression reduces to :

$L_{req}= Lt_{n-> \infty} ((2\pi n )^{1/2n})/em$ .

Let $L^{'} = Lt{n->\infty}(2\pi n )^{1/2n}$

$L{'}=e^{Lt_{n-> \infty} (ln(2\pi n)/2n}$ .

Thus by applying L' Hospital rule,

$L^{'}=1$ .

Therefore $L_{req} = \boxed{1/em}$ .