Know Your Limits

We know that $2\cos\left(\dfrac{\pi}{4}\right)=\sqrt{2}$ , and using repeatedly the identity $2\cos\left(\dfrac{\theta}{2}\right)=\sqrt{2+2\cos(\theta)}$ , we have $2\cos\left(\dfrac{\pi}{8}\right)=\sqrt{2+\sqrt{2}}$ , $2\cos\left(\dfrac{\pi}{16}\right)=\sqrt{2+\sqrt{2+\sqrt{2}}}$ , and in general, $2\cos\left(\dfrac{\pi}{2^n}\right)=\underbrace{\sqrt{2+\sqrt{2+\cdots}}}_{n-1 ~~ \text{roots}}$ .

Then, $L=2^n \underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\cdots}}}}_{n ~~ \text{roots}}=2^n \sqrt{2-\underbrace{\sqrt{2+\sqrt{2+\cdots}}}_{n-1 ~~ \text{roots}}}=2^n \sqrt{2-2\cos\left(\dfrac{\pi}{2^n}\right)}=2^n \cdot 2\sin\left(\dfrac{\pi}{2^{n+1}}\right)=2^{n+1}\sin\left(\dfrac{\pi}{2^{n+1}}\right)$ .

Finally, $\displaystyle \lim_{n \to \infty} L = \boxed{\pi}$ .

Put root2 = 2cospi/4.