How do I compute the following limit: $ \lim_{x \to \infty} \frac{x!}{\left( \frac{x}{e} \right)^{x}}$? In class we proved that $$ \lim_{x \to \infty} \frac{x!}{2^{x}} = \infty$$
This got me thinking for what value $n$ $$ \lim_{x \to \infty} \frac{x!}{n^{x}}$$ would the limit be $= 0$.
So clearly $n = x$ makes the bottom part of the fraction go to infinity much faster than the top part, and this is the case for $n = \frac{x}{2}$ as well. However, the limit for $n = \frac{x}{3}$ is $\infty$. I immediately became suspicious that the "turning point" would be for $n = \frac{x}{e}$. Due to calculator approximation errors, a normal TI-89 says the limit is $\infty$, but I'm not really sure if that's correct. 
In any case, how would one compute the limit for when $n = \frac{x}{e}$?
 A: Hint
When you have to manipulate factorials, Stirling approximation is very often the trick to be used.
As a first approximation, you have $$x!\approx \sqrt{2\pi\, x}\,\Big(\frac x e\Big)^x$$ which than makes for your problem $$\frac{x!}{\left( \frac{x}{e} \right)^{x}}\approx \sqrt{2\pi\,x} $$ Please, remember it : it is very useful and you will often need it !
A: Maybe a nice way of solving your problem:
$$\lim_{x\to\infty}\frac{x!}{\left(\frac{x}{e}\right)^x}=\lim_{x\to\infty}\frac{e^xx!}{x^x}=\lim_{x\to\infty}e^xx^{-x}x!=$$
$$\lim_{x\to\infty}\exp\left[x-x\ln(x)+\ln(x!)\right]=\exp\left[\lim_{x\to\infty}\left(x-x\ln(x)+\ln(x!)\right)\right]=$$
$$\exp\left[\lim_{x\to\infty}\ln\left(e^xx^{-x}!\right)\right]=\infty$$
A: Let
$r(x)
=\frac{x!}{(x/e)^x}
=\frac{x!e^x}{x^x}
$.
Then
$\begin{array}\\
s(x)
&=\frac{r(x+1)}{r(x)}\\
&=\frac{\frac{(x+1)!e^{x+1}}{(x+1)^{x+1}}}{\frac{x!e^x}{x^x}}\\
&=\frac{(x+1)ex^x}{(x+1)^{x+1}}\\
&=\frac{ex^x}{(x+1)^{x}}\\
&=\frac{e}{(1+1/x)^{x}}\\
\text{so that}\\
\ln(s(x))
&=1-x\ln(1+1/x)\\
\end{array}
$
We now use the expansion
$z-z^2/2
<\ln(1+z)
< z-z^2/2+z^3/3
$
for $0 < z < 1$.
Then
$x\ln(1+1/x)
\lt x(1/x-1/(2x^2)+1/(3x^3))
=1-1/(2x)+1/(3x^2)
$
and
$x\ln(1+1/x)
\gt x(1/x-1/(2x^2))
=1-1/(2x)
$.
Therefore
$1/(2x)-1/(3x^2)
\lt s(x)
\lt 1/(2x)
$.
This implies that
$\sum_{x=n}^m \ln(s(x))$
diverges as
$m \to \infty$.
But
$\begin{array}\\
\sum_{x=n}^m \ln(s(x))
&=\sum_{x=n}^m \ln(r(x+1)/r(x))\\
&=\sum_{x=n}^m (\ln(r(x+1))-\ln(r(x)))\\
&=\ln(r(m+1))-\ln(r(n))\\
\end{array}
$
and
this implies that
$\ln(r(m+1))$
diverges to $\infty$.
If we look more closely at this,
the argument shows that,
since
$\begin{array}\\
\sum_{x=n}^m \ln(s(x))
&\sim \sum_{x=n}^m 1/(2x)\\
&\sim \ln(m)/2\\
&=\ln(\sqrt{m})\\
\end{array}
$
that
$\ln(r(x))
\sim \ln(\sqrt{x})
$.
As is often the case,
nothing here is original.
