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 

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 $\displaystyle \ln \left( \frac{( \frac{n}{e} )^n}{ (\frac{(1-2\alpha)n}{e})^{(1-2\alpha)n}\,(\frac{2\alpha n}{e})^{2\alpha n}}\right)^{\frac{1}{n}}$:
$$\left( \frac{( \frac{n}{e} )^n}{ (\frac{(1-2\alpha)n}{e}  )^{(1-2\alpha)n}\,(\frac{2\alpha n}{e})^{2\alpha n}}\right)^{\frac{1}{n}}=\frac{\frac{n}{e}}{ (\frac{(1-2\alpha)n}{e}  )^{1-2\alpha}\,(\frac{2\alpha n}{e})^{2\alpha}}=\frac{1}{ (1-2\alpha)^{1-2\alpha}(2\alpha)^{2\alpha}}$$
