Showing that $x^x/(2x)!=0$ as $x$ approaches infinity

Question

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.

Answer 1

\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}

Answer 2

Note that $$ \frac{n^n}{(2n)!}=\frac{\overbrace{n\cdot n\cdot\ldots \cdot n}^{n \text{ factors}}}{n!\cdot\underbrace{(n+1)(n+2)\ldots(2n)}_{n\text{ factors}}}\le \frac1{n!}$$

Answer 3

Assuming $x$ is an integer in your question (so that I'll use $n$ instead of $x$, for the sake of my own ease of mind): a simple way, which uses a big (and quite overkill) "hammer," is to invoke Stirling's approximation:

$$ (2n)! \operatorname*{\sim}_{n\to\infty} 2\sqrt{\pi n}\frac{(2n)^{2n}}{e^{2n}} $$ and look at the limit (when $n\to\infty$) of $$ \frac{1}{2\sqrt{\pi n}}\cdot\frac{n^n e^{2n}}{(2n)^{2n}} = \frac{1}{2\sqrt{\pi n}}\cdot\frac{e^{2n}}{2^{2n}n^n} = \frac{1}{2\sqrt{\pi n}}\cdot\left(\frac{e^{2}}{4n}\right)^n $$