Show that $\lim_{n\rightarrow \infty} \int_0^{\pi/2} 2^n \sqrt{n} \sin^n(x) \cos^{n-2}(x) \; dx = \sqrt{2\pi}$ I wish to show
$$\lim_{n\rightarrow \infty} \int_0^{\pi/2} 2^n \sqrt{n} \sin^n(x) \cos^{n-2}(x) \; dx = \sqrt{2\pi}$$
I've tried substitution and integration by parts to get a recursive formula for the integral, but so far it hasn't worked. Any help would be appreciated, thanks.

I will present you the main steps to a final solution and the details on each step can be your job. 
First notice that a simple substiution $n=2k$ gives 
$$\lim_{n\rightarrow \infty} \int_0^{\pi/2} 2^n \sqrt{n} \sin^n(x) \cos^{n-2}(x) \; dx = \lim_{k\rightarrow \infty} \int_0^{\pi/2} 2^{2k} \sqrt{2k} \sin^{2k}(x) \cos^{2k-2}(x) \; dx$$ 
Now elaborating with the new expression in the following way we have 
$$2^{2k-1} \sqrt{2k}\cdot2\int_0^{\pi/2}\sin^{2(k+1/2)-1}(x) \cos^{2(k-1/2)-1}(x) \; dx$$
\begin{eqnarray}
&=& 2^{2k-1}\sqrt{2k}\cdot\beta(k+1/2,k-1/2) \\ 
&=& 2^{2k-1}\sqrt{2k}\frac{\Gamma(k+1/2)\Gamma(k-1/2)}{\Gamma(2k)} \\ 
&=& 2^{2k}\sqrt{2k}\frac{2k}{2k-1}\cdot\frac{\Gamma(k+1/2)^2}{\Gamma(2k+1)} \\ 
&=& 2^{2k}\sqrt{2k}\frac{2k}{2k-1}\cdot\frac{(\frac{(2k-1)!!\sqrt{\pi}}{2^{k}})^{2}}{(2k)!} \\
&=& \pi\cdot\sqrt{2k}\frac{2k}{2k-1}\cdot\frac{(2k-1)!!}{(2k)!!} \\
&=& \pi\sqrt{\frac{2k}{2k+1}}\frac{2k}{2k-1}\cdot\sqrt{\frac{(2k-1)!!^2(2k+1)}{(2k)!!^2}} \\ 
&=& \;\pi\sqrt{\frac{2k}{2k+1}}\frac{2k}{2k-1}\sqrt{\prod_{j=1}^{k}\frac{(2j-1)}{(2j)}\frac{(2j+1)}{(2j)}} \\ 
& \rightarrow & \pi\cdot\sqrt{1}\cdot1\sqrt{\prod_{j=1}^{\infty}\frac{(2i-1)}{(2i)}\frac{(2i+1)}{(2i)}} \\ 
&=&\pi\cdot\sqrt{\frac{2}{\pi}} \\ 
&=& \sqrt{2\pi},\;k\rightarrow \infty
\end{eqnarray}
