<s>[INST] How to calculate factorial function as $x\to\infty$? I need to calculate 
$$\lim_{x \to \infty} \frac{((2x)!)^4}{(4x)! ((x+5)!)^2 ((x-5)!)^2}.$$
Even I used Striling Approximation and Wolfram Alpha, they do not help.
How can I calculate this?
My expectation of the output is about $0.07$.
Thank you in advance.
 [/INST] The limit as given in the problem is equal to zero. This is shown by the following. 
Using $\Gamma(1+x) = x \Gamma(x)$ the expression to evaluate is seen as
\begin{align}
\phi_{n} &= \frac{\Gamma^{4}(2n+1) }{ \Gamma(4n+1) \Gamma^{2}(n+6) \Gamma(n-4)} \\
&= \frac{n^{2}(n-1)^{2}(n-2)^{2}(n-3)^{2}(n-4)^{2}}{(n+1)^{2}(n+2)^{2}(n+3)^{2}(n+4)^{2}(n+5)^{2}} \ \frac{\Gamma^{4}(2n+1)}{\Gamma(4n+1) \Gamma^{4}(n+1)}.
\end{align}
Now, by using Stirling's approximation, namely,
\begin{align}
\Gamma(n+1) \approx \sqrt{2 \pi} \ n^{n+1/2} \ e^{-n}
\end{align}
this expression becomes
\begin{align}
\phi_{n} &= \frac{\left(1-\frac{1}{n}\right)^{2}\left(1-\frac{2}{n}\right)^{2}
\left(1-\frac{3}{n}\right)^{2}\left(1-\frac{4}{n}\right)^{2}}{
\left(1+\frac{1}{n}\right)^{2} \left(1+\frac{2}{n}\right)^{2} \left(1+\frac{3}{n}\right)^{2} \left(1+\frac{4}{n}\right)^{2} \left(1+\frac{5}{n}\right)^{2}} \ \sqrt{ \frac{2}{\pi n} }
\end{align} 
Taking the limit as $n \rightarrow \infty$ leads to
\begin{align}
\lim_{n \rightarrow \infty} \frac{\Gamma^{4}(2n+1) }{ \Gamma(4n+1) \Gamma^{2}(n+6) \Gamma(n-4)}  = 0.
\end{align}
</s>