Sounds familiar?

the expression inside the limit is equal to half of perimeter of regular n-gon.

as $n \to \infty$ , the limit of the shape is a circle with radius = 1, hence limit $= \frac{C}{2} = \frac{2\pi}{2} = \pi$

First off, the problem with the question is that it fails to mention that the angle in the cosine is actually in degrees, converting it to radians, the question can be stated as:

$\displaystyle\lim_{n \to \infty} \frac{n}{2}\times\sqrt {2 - 2 \cos \frac{2\pi}{n} }$

$=\displaystyle\lim_{n \to \infty} \frac{n}{2}\times\sqrt {2\Bigg(1 - \cos \frac{2\pi}{n}\Bigg)}$

Using $1-\cos(2\theta) = 2\sin^2(\theta)$ :

$=\displaystyle\lim_{n \to \infty} \frac{n}{2}\times\sqrt {4\sin^2\frac{\pi}{n}}$

$=\displaystyle\lim_{n \to \infty} \frac{n}{\not2}\times\not2\sin\frac{\pi}{n}$

Substituting $\frac{1}{n}=h$

$=\displaystyle\lim_{h \to 0^+} \frac{\sin(\pi h)}{h}$

$=\displaystyle\lim_{h \to 0^+} \frac{\sin(\pi h)}{(\pi h)}\times\pi$

Using $\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1$ :

= $\boxed{\pi}$

Similar solution with @Aryaman Maithani

\begin{aligned} L & = \lim_{n \to \infty} \frac n2 \sqrt{2-2\color{#3D99F6}\cos \frac {2\pi}n} & \small \color{#3D99F6} \text{Since }\cos (2\theta) = 1 - 2\sin^2 \theta \\ & = \lim_{n \to \infty} \frac n2 \sqrt{2-2\color{#3D99F6}\left(1-2\sin^2 \frac \pi n\right)} \\ & = \lim_{n \to \infty} \frac n2 \sqrt{4\sin^2 \frac \pi n} \\ & = \lim_{n \to \infty} \frac n \sin^2 \frac \pi n & \small \color{#3D99F6} \text{Multiply up and down by }\frac \pi n. \\ & = \lim_{n \to \infty} \frac {\pi \color{#3D99F6}\sin^2 \frac \pi n}{\color{#3D99F6}\frac \pi n} & \small \color{#3D99F6} \text{Note that }\lim_{x \to 0} \frac {\sin x}x = 1 \\ & = \boxed{\pi} \end{aligned}

Let $\displaystyle u=\frac{2}{n}$ , then

$\displaystyle \lim_{n\to\infty} \frac{n}{2}\times\sqrt{2-2\cos{\frac{2\pi}{n}}}=\sqrt{2}\times\lim_{u\to 0} \frac{\sqrt{1-\cos{\pi u}}}{u}$

Using $\displaystyle \sin^2{\theta}=1-\cos^2{\theta}=(1+\cos{\theta})(1-\cos{\theta})$ , we get

$\displaystyle \sqrt{2}\times\lim_{u\to 0} \frac{\sqrt{1-\cos{\pi u}}}{u}=\sqrt{2}\times\lim_{u\to 0} \frac{\sin{\pi u}}{u\sqrt{1+\cos{\pi u}}}=\sqrt{2}\times\lim_{u\to 0} \frac{\pi \cos{\pi u}}{1}\times\lim_{u\to 0}\frac{1}{\sqrt{1+\cos{\pi u}}}$

Finally

$\displaystyle \lim_{n\to\infty} \frac{n}{2}\times\sqrt{2-2\cos{\frac{2\pi}{n}}}=\sqrt{2}\times\pi\times\frac{1}{\sqrt{2}}=\pi$