Limit of a function including factorials $$ \lim \frac{(2x)!}{(x! \cdot 2^x)^2} $$
How can I deal with problems including factorials as the same as this problem
 A: Just play around with the factorials by remembering what they mean
Rewrite as $$\lim_{x\rightarrow \infty}\frac{(2x)!}{(x!*2^x)(x!*2^x)}$$ Now notice that $(2x)!$ is the product of al the integers below $2x$. We could write that as $(2x)_x * x!$ where $(n)_x$ means the product of all integers below $n$ and above $x$. Then $x!$ cancels with the denominator and we get
$$\frac{(2x)_x}{2^x(x!*2^x)}$$
Now notice this: $$x!*2^x=(x*(x-1)*\cdots*2*1) * (2*2*\cdots*2*2)$$
Rearrange these factors(put a $2$ next to each factor of $x!$):
$$x!*2^x=(2x)*(2(x-1)*\cdots*(2*2)*(2*1))\leq(2x)_x$$
Which cancels again. We are left with
$$\lim_{x\rightarrow \infty}\frac{(2x)!}{(x!*2^x)^2}\leq\lim_{x\rightarrow\infty} \frac{1}{2^x} = 0$$
So
$$\lim_{x\rightarrow \infty}\frac{(2x)!}{(x!*2^x)^2} = 0 $$
(since it can't be negative)
A: Note that as x tends to $\infty$, we can use the Stirling Approximation.
That is, for large x, we have 
$$x!\approx\sqrt{2\pi x}\left(\frac{x}{e}\right)^x$$
We get 
$$\lim_{x \to \infty} \frac{2 \sqrt{\pi x}(\frac{2x}{e})^{2x}}{2\pi x(\frac{x}{e})^{2x}\cdot2^{2x}}$$
$$=\lim_{x \to \infty} \frac{1}{\sqrt {\pi x}}$$
$$=0$$
