Binomials and Limits

The central binomial coefficient has the following asymptotic property:

$n\to\infty\implies \binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}}$

This result comes directly from Stirling's approximation as follows:

$n\to\infty \implies n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\\ \implies\binom{2n}{n}=\frac{(2n)!}{(n!)^2}\sim~\frac{\sqrt{4\pi n}\left(\dfrac{2n}{e}\right)^{2n}}{\left\{\sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n\right\}^2}=\frac{4^n\cdot 2\sqrt{\pi n}\left(\dfrac{n}{e}\right)^{2n}}{2\pi n\left(\dfrac{n}{e}\right)^{2n}}=\frac{4^n}{\sqrt{\pi n}}$

$\implies \binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}}\textrm{ as }n\to\infty$

Similarly, it can be obtained that,

$n\to\infty\implies \binom{4n}{2n}\sim \frac{4^{2n}}{\sqrt{2\pi n}}$

Use these approximations to rewrite the original limit as,

$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\left(\frac{4^n\cdot 4^n\cdot \sqrt{2\pi n}}{4^{2n}\cdot \sqrt{\pi n}}\right)=\sqrt2$