<s>[INST] Showing that $x^x/(2x)!=0$ as $x$ approaches infinity How would I show
$$\lim_{x\to\infty} \frac{x^x}{(2x)!}=0$$
I know $x^x$ grows faster than $(2x)!$
So then would I do 
$$\frac{x^x}{2x(2x-1)(2x-2)(2x-3)\cdots(2x-(2x-1))}$$
But how do I proceed.
 [/INST] \begin{align}
\frac{x^x}{(2x)!} & = \frac{\overbrace{x\cdots x}^{x\text{ factors}}}{\underbrace{1\cdot2\cdot3\cdots x}_{x\text{ factors}} \cdot \underbrace{(x+1) \cdot (x+2) \cdots (2x)}_{x\text{ factors}}} \\[10pt]
& = \frac 1 {x!} \cdot \underbrace{\frac x {x+1} \cdot \frac x {x+2} \cdot \frac x {x+3} \cdots \frac x {x+x}}_\text{This is $<1$.} \\[10pt]
& < \frac 1 {x!} \to 0 \text{ as } x\to\infty. 
\end{align}
</s>