Finding limit of term involving binomial coefficient I am required to find the limit of
$$
\frac{\sqrt{n}}{2^n}\cdot \binom{n}{\frac{n}{2} + \sqrt{n}}
$$
I am given the following hints:

*

*Stirling's approximation
$$
n! \approx \sqrt{2\pi n}\cdot\left(\frac{n}{e}\right)^n
$$


*For $k \rightarrow \infty$, with $x$ being a constant.
$$
\left(1 + \frac{x}{k} \right)^k \rightarrow e^x
$$
I've tried doing a direct substitution using Stirling's approximation, but I am stuck with how to further simplify the terms. I am also unsure of how to use (2).
This is the expression I am currently stuck at and have no idea how to proceed:
$$
\frac{n^{n+1}}{2^n \cdot \sqrt{2\pi}}\frac{1}{\sqrt{(\frac{n}{2})^2-n}\cdot(\frac{n}{2}-\sqrt{n})^{\frac{n}{2}-\sqrt{n}}\cdot(\frac{n}{2}+\sqrt{n})^{\frac{n}{2}+\sqrt{n}}}
$$
Any hints is greatly appreciated.

Ignoring the $\sqrt{2\pi}$ factor, you can collect terms as follows: $$2 \cdot\left( \frac{n}{2\sqrt{n^2/4 -n}} \right)^{n+1} \cdot \left( \frac{n/2-\sqrt n}{n/2 +\sqrt n} \right)^{\sqrt n} $$ Now, the first factor is further simplified to $$\left( \frac{\sqrt n}{\sqrt {n-4}} \right)^{n+1} \\ = \frac{1}{ (1-\frac 4n)^{\frac{n+1}{2}}}\\ = \frac{1}{((1-\frac 4n)^n)^{1/2} \cdot (1-\frac 4n)^{1/2}} $$
$$ \to \frac{1}{(e^{-4})^{1/2}} \cdot \frac{1}{1}\\ = e^2 $$ and the second factor is $$\left(1-\frac{2\sqrt n}{n/2 +\sqrt n} \right)^{\sqrt n} \\ =\left( 1-\frac{2}{\sqrt n/2 +1} \right)^{\sqrt n} \\= \left( \left( 1-\frac{2}{\sqrt n/2 +1} \right)^{\sqrt n /2 +1} \right)^2 \cdot \left( 1-\frac{2}{\sqrt n/2 +1} \right)^{-2} \\ \to (e^{-2})^2 \cdot 1 \\ =e^{-4} $$
So, all in all the answer should be $$\frac{1}{\sqrt{2\pi}} \times 2 \times e^2 \times e^{-4}=\color{red}{ \sqrt{\frac{2}{\pi}} e^{-2}} $$
