How can we prove that $$\lim_{n\rightarrow\infty}\frac{\dbinom{2n}{n}^2}{\dbinom{4n}{2n}}=0?$$
From plotting this for some values of $n$ it is clear, but using an approximation like $\binom{n}{k}\approx \left(\frac{ne}{k}\right)^k$ gives a value of $1$, since both numerator and denominator are $(2e)^{2n}$.
Using Stirling's formula, one can show that $$ {2n\choose n}\sim \frac{2^{2n}}{\sqrt{\pi n}}$$ as $n\to\infty$, and therefore $$ \frac{{2n\choose n}^2}{{4n\choose 2n}}\sim\frac{\frac{2^{4n}}{\pi n}}{\frac{2^{4n}}{\sqrt{2\pi n}}}=\frac{\sqrt{2\pi n}}{\pi n}=\sqrt{\frac{2}{\pi n}}$$ Therefore $\frac{{2n\choose n}^2}{{4n\choose 2n}}\to 0$ as $n\to\infty$.
$$ \binom{2n}{n}= \frac{2^n (2n-1)!!}{n!} = 4^n \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)\tag{1}$$ hence: $$\begin{eqnarray*} \binom{2n}{n}^2\binom{4n}{2n}^{-1}&=&\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^2 \prod_{k=1}^{2n}\left(1-\frac{1}{2k}\right)^{-1}\\&=&\frac{1}{2n}\prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right)\prod_{k=2}^{2n}\left(1-\frac{1}{2k}\right)^{-1}\\&=&\frac{1}{2\sqrt{n}}\prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right)\prod_{k=2}^{2n}\left(1-\frac{1}{(2k-1)^2}\right)^{-1/2}.\tag{2}\end{eqnarray*}$$ Since $\prod_{k\geq 2}\left(1-\frac{1}{(2k-1)^2}\right)=\frac{\pi}{4}$ by the Weierstrass product for the cosine function, the product $\binom{2n}{n}^2\binom{4n}{2n}^{-1}$ behaves like $\frac{K}{\sqrt{n}}$ for large values of $n$ and the given limit equals zero.
