TITLE: Taking Limits with Binomial Coefficients
QUESTION [5 upvotes]: I am interested in taking the following limit:
\begin{equation}
\lim_{n \to \infty}\frac{{n/2 \choose m}}{n \choose m}.
\end{equation}
Provided that $m$ is fixed the solution is:
\begin{equation}
\lim_{n \to \infty}\frac{{n/2 \choose m}}{n \choose m}=2^{-m}.
\end{equation}
This can be seen by writing:
\begin{aligned}
\frac{{n/2 \choose m}}{n \choose m}&=\frac{\left(\frac{n}{2}\right)!(n-m)!}{\left(\frac{n}{2}-m\right)!n!}\\
&=\prod_{j=0}^{m-1}\frac{\frac{n}{2} -j}{n-j}\\
&=\frac{1}{2^m}\prod_{j=1}^{m-1}\frac{n-2j}{n-j}
\end{aligned}
and letting $n$ tend to infinity.
What however, can be said when $m$ depends on $n$? For example, if $m=\sqrt{n}$? Do we still have the quantity behaving asymptotically like $2^{-\sqrt{n}}$?

REPLY [2 votes]: Rewrite
$$
\lim_{n \to \infty}\frac{\dbinom{\frac{n}{2}}{m}}{\dbinom{n}{m}}=\lim_{n \to \infty}\frac{\left(\frac{n}{2}\right)!}{\left(\frac{n}{2}-m\right)!\ m!}\cdot\frac{(n-m)!\ m!}{n!}=\lim_{n \to \infty}\frac{\left(\frac{n}{2}\right)!\ (n-m)!}{n!\ \left(\frac{n}{2}-m\right)!}.\tag1
$$
Since $n\to\infty$, we can use Stirling's formula:
$$
k!\sim\sqrt{2\pi k}\left(\frac{k}{e}\right)^k\quad\text{as}\quad k\to\infty.
$$
Thus, we can use Stirling's formula to obtain the limit in $(1)$. I hope this help.
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