Evaluating $\lim_{n \to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$ I was thinking about different ways of finding $\pi$ and stumbled upon what I'm sure is a very old method: dividing a circle of radius $r$ up into $n$ isosceles triangles each with radial side length $r$ and central angle $$\theta=\frac{360^\circ}{n}$$ Use $s$ for the side opposite to $\theta$.

Then we can approximate the circumference as $sn$. By the law of cosines: $$s=\sqrt{2r^2-2r^2 \cos{\theta}}=r\sqrt{2-2\cos{\left(\frac{360^\circ}{n}\right)}}$$
We know $\pi=\text{circumference}/\text{diameter} \approx \frac{sn}{2r}=\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$. This becomes exact at the limit:
$$\pi=\lim_{n \to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$$
Now for my question: How would you solve the opposite problem? To make my meaning more clear, above I used the definition of $\pi$ to determine a limit that gives its value. But if I had just been given the limit, what technique(s) could I have used to determine that it evaluates to $\pi$?
 A: $$\sqrt{2-2\cos x}=2\left|\sin\frac{x}{2}\right|$$
hence everything boils down to:
$$ \lim_{x\to 0}\frac{\sin x}{x}=1.$$
A: Notice, $$\lim_{n\to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{2\pi}{n}\right)}$$
$$=\lim_{n\to \infty}\frac{n}{2}\sqrt{2\left(1-\cos\left(\frac{2\pi}{n}\right)\right)}$$ $$=\lim_{n\to \infty}\frac{n}{2}\sqrt{2\left(2\sin^2\left(\frac{\pi}{n}\right)\right)}$$ $$=\lim_{n\to \infty}\frac{2n}{2}\sin\left(\frac{\pi}{n}\right)$$ $$=\lim_{n\to \infty}\frac{\sin\left(\frac{\pi}{n}\right)}{\frac{1}{n}}$$ $$=\pi\lim_{n\to \infty}\frac{\sin\left(\frac{\pi}{n}\right)}{\left(\frac{\pi}{n}\right)}$$ Let $\frac{\pi}{n}=t\implies t\to 0\ as\ n\to \infty$ $$=\pi \lim_{t\to 0}\frac{\sin t}{t}$$ $$=\pi\times 1=\pi$$
A: Since there were already two solid answers that provided efficient approaches, I thought that it would be instructive to see a different way forward.  
Here, we will use the expansion of the cosine as 
$$\cos x=1-\frac12 x^2+O(x^4) \tag 1$$
Letting  $x=\frac{2\pi}{n}$ in $(1)$ yields  
$$2-2\cos \left(\frac{2\pi}{n}\right)=\frac{4\pi^2}{n^2} +O(n^{-4})$$
Finally, we have
$$\begin{align}
\frac{n}{2}\sqrt{2-2\cos \left(\frac{2\pi}{n}\right)}&=\frac n2 \sqrt{\frac{4\pi^2}{n^2} +O(n^{-4})}\\\\
&=\frac n2 \frac{2\pi}{n}\left(1+O(n^{-2})\right)\\\\
&=\pi+O(n^{-3})\to \pi
\end{align}$$
as expected!
A: 
Just recall double angle trig. identity: $\cos 2x=1-2\sin^2x$,

Starting from,
$$\begin{align}
\lim_{n\to \infty}\frac{n}{2}\sqrt{2-2\cos\left(2\pi/n\right)}\\=\lim_{n\to \infty}\frac{n}{2}\sqrt{2-2+4\sin^2\left(\pi/n\right)}\\
=\lim_{n\to \infty}\frac{n}{2}\cdot 2\sin\left(\pi/n\right)\\
=\pi\cdot \lim_{n\to \infty}\frac{\sin(\pi/n)}{\pi/n}\\
=\pi\cdot \lim_{t\to 0}\frac{\sin(t)}{t}\\
=\pi\cdot 1\\=\pi
\end{align}$$
