How to compute this limit $\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}$ $$\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}=\frac{3}{2}((1-2\alpha) \log{2\alpha}+2\alpha\log2\alpha)$$
such that  $2\alpha n\le n$
I tried to use Stirling formula and we get 
$$\lim_{n\to ∞}\frac{1}{n}\log{{n\choose 2\alpha n}}=\lim_{n\to ∞}\frac{1}{n}\log\frac{n^{\frac{3n}{2}}}{2\pi(n-2\alpha n)^{\frac{3((n-2\alpha n)}{2}{(2\alpha n)}^{3\alpha n}}}=$$
$$=\lim_{n\to ∞}\log{\frac{n^{\frac{3}{2}}}{2\pi(n-2\alpha n)^{\frac{3((1-2\alpha )}{2}{(2\alpha n)}^{3\alpha }}}}$$
but I couldn't continue 
 A: Assuming that $$\binom{m}{n} = \frac{\Gamma(m + 1)}{\Gamma(n + 1)\Gamma(m - n + 1)}\tag{1}$$ whenever the RHS is defined we can see that the sequence $$a_{n} = \binom{n}{2\alpha n}$$ satisfies $$\frac{a_{n + 1}}{a_{n}} = \frac{\Gamma(n + 2)}{\Gamma(2\alpha n + 2\alpha + 1)\Gamma(n + 2 - 2\alpha n - 2\alpha)}\cdot\frac{\Gamma(2\alpha n + 1)\Gamma(n - 2\alpha n + 1)}{\Gamma(n + 1)}$$ so that $$\frac{a_{n + 1}}{a_{n}} = \frac{n\Gamma(2\alpha n)(n - 2\alpha n)\Gamma(n - 2\alpha n)}{\Gamma(2\alpha n + 2\alpha)(n + 1 - 2\alpha n - 2\alpha)(n - 2\alpha n - 2\alpha)\Gamma(n - 2\alpha n - 2\alpha)}$$ and hence
\begin{align}
L &= \lim_{n \to \infty}\frac{a_{n + 1}}{a_{n}}\notag\\
&= \frac{1}{1 - 2\alpha}\lim_{n \to \infty}\frac{\Gamma(2\alpha n)\Gamma(n - 2\alpha n)}{\Gamma(2\alpha n + 2\alpha)\Gamma(n - 2\alpha n - 2\alpha)}\notag\\
&= \frac{1}{1 - 2\alpha}\lim_{n \to \infty}\frac{(2\alpha n/e)^{2\alpha n}((n - 2\alpha n)/e)^{n - 2\alpha n}}{((2\alpha n + 2\alpha)/e)^{2\alpha n + 2\alpha}((n - 2\alpha n - 2\alpha)/e)^{n - 2\alpha n - 2\alpha}}\notag\\
&= \frac{1}{1 - 2\alpha}\lim_{n \to \infty}\frac{(2\alpha n)^{2\alpha n}(n - 2\alpha n)^{n - 2\alpha n}}{(2\alpha n + 2\alpha)^{2\alpha n + 2\alpha}(n - 2\alpha n - 2\alpha))^{n - 2\alpha n - 2\alpha}}\notag\\
&= \frac{(2\alpha)^{-2\alpha}}{1 - 2\alpha}\lim_{n \to \infty}\left(\frac{n}{n + 1}\right)^{2\alpha n}\frac{1}{(n + 1)^{2\alpha}}\frac{(n - 2\alpha n)^{n - 2\alpha n}}{(n - 2\alpha n - 2\alpha))^{n - 2\alpha n - 2\alpha}}\notag\\
&= \frac{(2\alpha e)^{-2\alpha}}{1 - 2\alpha}\lim_{n \to \infty}\left(\frac{n - 2\alpha n - 2\alpha}{n + 1}\right)^{2\alpha}\left(\frac{n - 2\alpha n}{n - 2\alpha n - 2\alpha}\right)^{n(1 - 2\alpha)}\notag\\
&= (2\alpha e)^{-2\alpha}(1 - 2\alpha)^{2\alpha - 1}\lim_{n \to \infty}\left(1 - \frac{2\alpha}{n(1 - 2\alpha)}\right)^{n(2\alpha - 1)}\notag\\
&= (2\alpha)^{-2\alpha}(1 - 2\alpha)^{2\alpha - 1}\notag
\end{align}
and hence $a_{n}^{1/n}$ tends to the same limit $L = (2\alpha)^{-2\alpha}(1 - 2\alpha)^{2\alpha - 1}$ and therefore $$\frac{1}{n}\log a_{n} \to \log L = -\{(1 - 2\alpha)\log(1 - 2\alpha) + 2\alpha \log 2\alpha\}$$ We have used the Stirling's approximation $$\lim_{x \to \infty}\frac{\Gamma(x)}{(x/e)^{x}\sqrt{2\pi/x}} = 1$$ and the result makes sense only when $0 < \alpha < 1/2$. The terms with $\sqrt{2\pi/\cdots}$ are not seen in above limit evaluation because after some cancellation in numerator and denominator they tend to $1$ as $n \to \infty$.
Silly Note: A lot of typing in $\mathrm\LaTeX$ could be simplified by replacing $2\alpha$ with $a$ but alas! I chose to be in sync with OP's notation.
A: The Stirling formula is the right way. But ignore the logarithm first. I get 
$$\ln \frac{1}{ (1-2\alpha)^{1-2\alpha}(2\alpha)^{2\alpha}}$$
It comes from $\displaystyle \ln \left( \frac{( \frac{n}{e} )^n}{ (\frac{(1-2\alpha)n}{e})^{(1-2\alpha)n}\,(\frac{2\alpha n}{e})^{2\alpha n}}\right)^{\frac{1}{n}}$:
$$\left( \frac{( \frac{n}{e} )^n}{ (\frac{(1-2\alpha)n}{e}  )^{(1-2\alpha)n}\,(\frac{2\alpha n}{e})^{2\alpha n}}\right)^{\frac{1}{n}}=\frac{\frac{n}{e}}{ (\frac{(1-2\alpha)n}{e}  )^{1-2\alpha}\,(\frac{2\alpha n}{e})^{2\alpha}}=\frac{1}{ (1-2\alpha)^{1-2\alpha}(2\alpha)^{2\alpha}}$$
A: You could try instead an approximation for binomial coefficient: $\binom{n}{t} \sim \frac{n^t}{t!}$.
EDIT: OK I actually used Stirling's for the second part, this is $\log t! \sim 2 \alpha n \log 2 \alpha n +\frac{\log 4 \pi n}{2n} - 2 \alpha$, and then you divide through $n$ and get rid of $2 \alpha \log n $to get the final result. 
A: As I wrote in a comment, there is a problem somewhere.
Considering $$\binom{n}{2 a n}=\frac{n!}{(2an)!\,(n-2an)!}$$ $$\log (\binom{n}{2 a n})=\log(n!)-\log((2an)!)-\log((n-2an)!)$$ and using Stirling approximation of $\log(p!)$ for each of the terms, we should find, for infinitely large values of $n$ $$\frac 1 n\,\log (\binom{n}{2 a n})=-(1-2 a) \log (1-2 a)-2 a \log (2 a)+O\left(\frac{1}{n}\right)$$ which is what @user90369 found too.
