How can I compute this limit? I know that $\lim_{n\to\infty}n\sqrt{2-2\cos(\frac{2\pi}{x})}=2\pi$ but I am unsure about how to compute it by hand. 
I the limit of $n$ alone is infinity, obviously, and the limit of $\cos(\frac{2\pi}{x})=1$. Therefore, to me, is it not $n*\sqrt{2-2}=n*0=\infty*0=0?$

Hint: $\require{cancel}\;n\sqrt{2\left(1-\cos(\frac{2\pi}{n})\right)}=n\sqrt{2 \cdot 2 \sin^2(\frac{\bcancel{2} \pi}{\bcancel{2}n})}=2n\left|\sin\left(\frac{\pi}{n}\right)\right| = 2\pi \cdot \left|\cfrac{\sin(\frac{\pi}{n})}{\frac{\pi}{n}}\right|$
