Finding the limit: $\lim_{x\to \infty}\frac{1}{2}x\sin {\frac{180(x-2)}{x}}$ While investigating a problem, I came across a function: $$f(x) = \frac{1}{2}x\sin {\frac{180(x-2)}{x}}$$  When looking at the function in Desmos (I was checking my proof), I discovered that $$\lim_{x\to \infty}\frac{1}{2}x\sin {\frac{180(x-2)}{x}} = \pi$$ I double-checked Wolframalpha, and this limit is true. The only issue is that I can't seem to prove it by hand, and I'm really interested as to how $\pi$ pops out of nowhere. Please note I am working in degrees, so it is not 180 radians in the sin function. I would really appreciate it if someone could explain a solution.

By assumption that your $180$ it's $180^{\circ}$ we obtain: $$f(x)=\frac{1}{2}x\sin\frac{2\pi}{x}=\frac{\sin\frac{2\pi}{x}}{\frac{2\pi}{x}}\cdot\pi\rightarrow\pi.$$
