Showing that $\frac{x!}{x^{x}}$ tends to zero as x tends to infinity The question is pretty much in the title, I'm having difficulty formally showing that $\lim\limits _{x\to\infty}\frac{x!}{x^{x}}=0$ (despite intuitively it's fairly obvious).

Thanks in advance for helping ;)
We have that 

$$n! \leq \frac{n^n}{2^n} \text{ for } n \geq 6$$

so 

$$\lim_{n \to \infty} \frac{n!}{n^n} \leq \lim_{n\to \infty} \frac{1}{2^n} = 0.$$

For $x \in \mathbb{R}$, we know that $\Gamma$ function is increasing for inputs $\geq 2$, so the limit still holds (assume $x &gt; n \geq 6$):

\begin{align}
\Gamma(n) &amp;= (n-1)! \\
\Gamma(x) &amp;\leq \Gamma(\lfloor x+1 \rfloor) = \lfloor x \rfloor! 
\end{align}

so

$$\lim_{x \to \infty} \frac{\Gamma(x+1)}{x^x} \leq
\lim_{x \to \infty} \frac{\lfloor x+1 \rfloor!}{\lfloor x \rfloor^{\lfloor x \rfloor}} = 
\lim_{n \to \infty} \frac{(n+1)!}{n^n} \leq 
\lim_{n\to \infty} \frac{n+1}{2^n} = 0.$$

I hope this helps ;-)

Edit: Fixed some minor issues, thanks to @julien for noticing.
Hint: What is $\log(n!)$? Then replace a sum with integrals