$$\begin{align}\frac{(2n)!}{2^{2n}(n!)^2}&=\frac{(2n)^{2n}e^{-2n}\sqrt{2\pi\times2n}(1+\varepsilon_n)}{2^{2n}(n^ne^{-n}\sqrt{2\pi n}(1+\varepsilon_n))}\\ &=\frac{\cancel{2^{2n}}\cancel{n^{2n}}\cancel{e^{-2n}}\cancel2\sqrt{\pi n}(1+\varepsilon_n)}{\cancel{2^{2n}}\cancel{n^{2n}}\cancel{e^{-2n}}\times\cancel2\pi n(1+\varepsilon_n)}\\ &=\frac1{\sqrt{\pi n}}\times\frac{1+\varepsilon_{2n}}{(1+\varepsilon_n)^2}\\ &=\frac1{\sqrt{\pi n}}(1+\underbrace{\beta_n}_{{\underset{n\to+\infty}\longrightarrow}0})\\ &\sim\frac1{\sqrt{\pi n}}\end{align}$$