Calculate limit with factorial I need to find the limit of this function..I thought about L'hôpital's rule, but can't seem to derive them both..
$$\lim_{n\rightarrow\infty} \frac{(2n)!}{(n!)^2}$$
 A: Hint: Note that 
$$\frac{(2n)!}{n!n!}=\frac{(2n)(2n-1)(2n-2)\cdots (n+1)}{(n)(n-1)(n-2)\cdots (1)}.$$
Each of the $n$ fractions $\frac{2n}{n}$, $\frac{2n-1}{n-1}$, $\frac{2n-2}{n-2}$, and so on down to $\frac{n+1}{1}$ is $\ge 2$.  So the thing blows up quite fast.
A: Let us assume that you use Stirling approximation which is $$p! \simeq \sqrt {2 \pi p} \Big(\frac{p}{e}\Big)^p$$ So, applying it to your terms, you get $$I_n=\frac{(2n)!}{(n!)^2}=\frac{2^{2 n}}{\sqrt{\pi n}} $$ which, as nicely demonstrated by André Nicolas in his answer, increases very fast (if $n=10$, the exact value of the expression is $184756$ while the approximation I used gives $187079$).
This solution is probably off-topic to you but just remember this approximation of $p!$; for sure you will be using a lot in your studies.
Added later
There is even a simpler way using logarithms and Stirling first approximation $$\log(p!) \simeq p \log(p)-p$$ So, in your case $$\log(I_n) \simeq \Big(2n \log(2n)-2n\Big)-2 \Big(n \log(n)-n\Big)=2n\log(2)$$ So, $I_n \simeq2^{2n}$.
