Let $x_0=2\cos\frac{\pi}{6}$ and $x_n=\sqrt{2+x_{n-1}},n=1,2,3...,$ prove that $\lim_{n\to \infty}2^{n+1}\sqrt{2-x_n}=\frac{\pi}{3}$ Let $x_0=2\cos\frac{\pi}{6}$ and $x_n=\sqrt{2+x_{n-1}},n=1,2,3...,$ prove that $\lim_{n\to \infty}2^{n+1}\sqrt{2-x_n}=\frac{\pi}{3}$
I am not able to correctly solve it,made some attempts,but no luck.This $2^{n+1}$ is creating problem.How should i prove this limit to be $\frac{\pi}{3}$. 

As Suggested by @Did in comment,here is a solution:
First using induction on $n$ prove that $x_n=2 cos(\frac{\pi}{6 \cdot 2^n})$,then
From identity $\sin^2 \frac{t}{2}=\frac{1}{2}(1-\cos t)$ we have
\begin{align}
\lim_{n\rightarrow\infty}{2^{n+1}\sqrt{2-2\cos\left(\frac{\pi}{6 \cdot2^n}\right)}} & = \lim_{n\rightarrow \infty}{2^{n+2}\sin\left(\frac{\pi}{6\cdot 2^{n+1}}\right)} \\
& = \lim_{n\rightarrow \infty}\frac{\sin\left(\frac{\pi}{3 \cdot2^{n+2}}\right)}{\frac{1}{2^{n+2}}} \\
& = \lim_{t\rightarrow 0^+}\frac{\sin\left(\pi t/3\right)}{t} \\
& = \pi/3 \\
\end{align}
