Trigonometry Problem..

This is a basic limits problem. Set $t=\frac{2\pi}{x}$ , hence, the limit becomes t approaches 0. With some algebra, the problem reduces to $\frac{\pi\sin{t}}{ t}$ , which is $\pi * 1$ .

You can also write it as: $\frac{\sin{2\pi/x}}{\frac{2}{x}}$ and apply L'Hopital's rule.

$\begin{aligned} L & = \lim_{x \to \infty} \frac x2 \sin \frac {2\pi}x \\ & = \lim_{x \to \infty} \pi \cdot \frac {\sin \frac {2\pi}x}{\frac {2\pi}x} & \small \color{#3D99F6} \text{Let }u = \frac {2\pi}x \\ & = \lim_{u \to 0} \pi \cdot \frac {\sin u}u \\ & = \boxed \pi \end{aligned}$

Just multiply by $\frac{\pi}{\pi}$ and let $u = x/2\pi$

The use $t->0 => sent=t$