Evaluation of $\lim_{n\rightarrow \infty}\frac{n!\cdot e^n}{\sqrt{n}n^n}$ without stirling approximation 
Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\frac{n!\cdot e^n}{\sqrt{n}\cdot n^n}$

$\bf{My\; Try::}$we can write it as $$l=\lim_{n\rightarrow \infty}\frac{e^n}{\sqrt{n}}\cdot \left(\frac{1}{n}\cdot \frac{2}{n}\cdot \frac{3}{n}\cdots \cdots \frac{n}{n}\right)$$
$$\ln (l) = \lim_{n\rightarrow \infty}\bigg[n-\frac{1}{2}n+\sum^{n}_{r=1}\ln\left(\frac{r}{n}\right)\bigg]$$
Now how can i solve it, Help required, Thanks

Let
$$
a_n : = \frac{{n!e^n }}{{n^n \sqrt n }}
$$
and $a:= \lim a_n$. I assume that this limit exists and is finite and positive. Then
\begin{align*}
a_{2n} & = \frac{{(2n)!e^{2n} }}{{(2n)^{2n} \sqrt {2n} }} = \frac{{(2n)!}}{{n!^2 2^{2n} }}\frac{{(n!)^2 e^{2n} }}{{n^{2n + 1} }}\sqrt {\frac{n}{2}}  = \sqrt {\frac{{1 \cdot 1}}{{2 \cdot 2}}\frac{{3 \cdot 3}}{{4 \cdot 4}} \cdots \frac{{(2n - 1)(2n - 1)}}{{2n \cdot 2n}}} a_n^2 \sqrt {\frac{n}{2}} 
\\ & = \sqrt {\frac{{1 \cdot 3}}{{2 \cdot 2}}\frac{{3 \cdot 5}}{{4 \cdot 4}} \cdots \frac{{(2n - 1)(2n + 1)}}{{2n \cdot 2n}}} a_n^2 \sqrt {\frac{n}{{4n + 2}}} .
\end{align*}
Taking the limit of both sides gives
$$
a = \sqrt {\mathop {\lim }\limits_{n \to  + \infty } \frac{{1 \cdot 3}}{{2 \cdot 2}}\frac{{3 \cdot 5}}{{4 \cdot 4}} \cdots \frac{{(2n - 1)(2n + 1)}}{{2n \cdot 2n}}} a^2 \frac{1}{2}.
$$
The infinite product under the square root is the reciprocal of the famous Wallis product, whence
$$
a = \sqrt {\frac{2}{\pi }} a^2 \frac{1}{2},
$$
i.e., $a=\sqrt{2\pi}$.
