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?

Recalling the Stirling's approximation $$n!\sim\sqrt{2\pi}n^{n+1/2}e^{-n}
 $$ we have $$\sqrt{2\pi}\lim_{n\rightarrow\infty}\frac{n^{2n+1}e^{-2n}}{\left(2n\right)^{2n+1/2}e^{-2n}}=0.
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