$$\lim_{n\to\infty}\binom{n}{50}\left(\frac2n\right)^{50}\left(1-\frac2n\right)^{n-50}$$
Taking $nh=1$ and $K=\binom{n}{50}\left(\frac2n\right)^{50}\left(1-\frac2n\right)^{n-50}$, we have: $$\ln K =\ln\left[\binom{n}{50}(2h)^{50}\right]+(1-50h)\frac{\ln(1-2h)}h \\=\ln\left[\frac{2^{50}}{50!}\prod_{k=1}^{49}(1-kh)\right]+(1-50h)\frac{\ln(1-2h)}h$$ Now I think $$\lim_{n\to\infty}K=\frac{2^{50}}{50!}\frac1{e^2}$$ Is this correct way to solve this? Is my answer correct?
Edit: I noted that: $$\left(\frac2n+1-\frac2n\right)^n=\sum_{k=0}^{n}\underbrace{\binom{n}{k}\left(\frac2n\right)^{k}\left(1-\frac2n\right)^{n-k}}_{t_k}$$ As $n\to\infty$, LHS is exactly 1 and we need to find the minute contribution of $t_{50}$.
I would approach this using the two following hints.
Hint 1: $$ \binom{n}{50}\left(\frac2n\right)^{50} =\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\cdots\frac{n-49}{n}\frac{2^{50}}{50!}\\ $$ Hint 2: $$ \left(1-\frac2n\right)^{n-50} =\left(1-\frac2n\right)^n\left(1-\frac2n\right)^{-50}\\ $$