How do I prove that $\lim_{n\to\infty} n\sqrt{2-2\cos(\frac{2\pi}{n})}=2\pi$

Question

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?

Answer 1

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}

Answer 2

Hint: $\lim_{n\to\infty}n(f(\frac1n)-f(0))=\lim_{h\to0}\frac{f(h)-f(0)}{h}$

Answer 3

Hint: set $2\pi/n=2t$ and apply the bisection formula

Answer 4

take $2$ outside, write, $1 - \cos 2(x) = 2 \sin^2(x)$ and take it outside of sqrt. then use this.

Answer 5

Using power series:

$$\cos\frac{2\pi}n=\sum_{k=0}^\infty(-1)^n\frac{(2\pi)^{2n}}{n^{2n}(2n)!}\implies 2-2\cos\frac{2\pi}n=2\left(\left(\frac{2\pi}n\right)^2\frac12-\left(\frac{2\pi}n\right)^2\frac1{24}+\ldots\right)$$

and from here

$$n\sqrt{2-2\cos\frac{2\pi}n}=n\sqrt{\left(\frac{2\pi}n\right)^2+\mathcal O\left(\frac1{n^2}\right)}\xrightarrow[n\to\infty]{}2\pi$$