How do I prove that $\lim_{n\to\infty} n\sqrt{2-2\cos(\frac{2\pi}{n})}=2\pi$ I have to prove that
$$
\lim_{n\to\infty} n\sqrt{2-2\cos\left(\frac{2\pi}{n}\right)}=2\pi
$$
It's geometrically obvious since it is the limit of the $n$-gons inside a unit circle.
But how do I prove it?

From identity $\sin^2 \frac{t}{2}=\frac{1}{2}(1-\cos t)$ we have
\begin{align}
\lim_{n\rightarrow\infty}{n\sqrt{2-2\cos\left(\frac{2\pi}{n}\right)}} & = \lim_{n\rightarrow \infty}{2n\sin\left(\frac{\pi}{n}\right)} \\
& = 2\pi\lim_{n\rightarrow \infty}\frac{\sin\left(\frac{\pi}{n}\right)}{\frac{\pi}{n}} \\
& = 2\pi\lim_{t\rightarrow 0^+}\frac{\sin\left(t\right)}{t} \\
& = 2\pi(1) \\
& = 2\pi
\end{align}
