Evaluate $ \lim_{n\to\infty} \frac{(n!)^2}{(2n)!} $ I'm completely stuck evaluating $ \lim_{n\to\infty} \frac{(n!)^2}{(2n)!} $ how would I go about solving this?

As all terms are positive, we have $$0 \leq \frac{(n!)^2}{(2n)!} =  \frac{n!}{2n \cdot \dots \cdot (n+1)} = \prod_{k=1}^n \frac{k}{k+n} \leq \prod_{k=1}^n \frac{1}{2} = \left(\frac{1}{2}\right)^n$$
So then as 
$$\lim\limits_{n\rightarrow\infty} \left(\frac{1}{2}\right)^n = 0$$
It follows that
$$\lim\limits_{n\rightarrow\infty} \frac{(n!)^2}{(2n)!} = 0$$
Edit:
For what it's worth if you want quick justification of the second inequality step:
$$k \leq n \implies 2k \leq k + n \implies \frac{k}{k+n} \leq \frac{1}{2}$$
