The problem is to find the limit
$$\lim_{n\to\infty} \frac{n!e^n}{n^n}.$$
My first idea was reorder terms:
$$\lim_{n\to\infty}\frac{n!e^n}{n^n}=\lim_{n\to\infty} n!\left(\frac{e}{n}\right)^n$$
with the indeterminate form $\infty \cdot 0$. Reordering terms:
$$\lim_{n\to\infty} \frac {\left(\frac{e}{n}\right)^n}{1/n!}$$
with the form $0/0$. Can I apply L'Hopital's rule to evaluate the limit? Using WolframAlpha, the answer is that the limit don't exist ($\infty$):
https://www.wolframalpha.com/input?i=limit+x-%3E+inf+%28x%21e%5Ex%2Fx%5Ex%29,
but i can't find the way to find this answer!!
Best Answer
First of all, L'Hopital's rule is for some differentiable function. Sure, you can do it if you replace the factorial with the Gamma function, but this is a bit annoying.
Anyway, another way to do this: denote by $ a_n := \dfrac{n!e^n}{n^n} $. Then we have the ratio $$ \frac{a_{n+1}}{a_n} = \frac{e}{\left(1 + \frac{1}{n}\right)^n} .$$ Take logs and you get $$ \log{a_{n+1}} - \log{a_n} = 1 - n\log\left(1 + \frac{1}{n}\right) = \frac{1}{2n} + O\!\left(\frac{1}{n^2}\right)$$ using the Maclaurin expansion of $\log(1 + x)$.
We then have that $\log{a_n} \to \infty$ by telescoping and the divergence of the harmonic series, and so $a_n \to \infty$ as well.