limit to pi of $\sin(\frac{x+\pi}{x-\pi})\sin(\frac{x-\pi}{x+\pi})$

Question

Finding $\lim_{x\rightarrow\pi}{\sin(\frac{x+\pi}{x-\pi})\sin(\frac{x-\pi}{x+\pi})}$.

I'm thinking on the lines of squeeze theorem after I convert it into the seperate cosine form. but the numbers are just all weird.

Answer 1

Hint: First term in product oscillates between $[-1,1]$, while the second term goes to $0$.

Answer 2

$$-\sin(\frac{x-\pi}{x+\pi})\leq{\sin(\frac{x+\pi}{x-\pi})\sin(\frac{x-\pi}{x+\pi})}\leq \sin(\frac{x-\pi}{x+\pi})$$

since $-1\leq \sin(x)\leq1$

$$\lim_{x\to\pi}(-\sin(\frac{x-\pi}{x+\pi}))=\lim_{x\to\pi}(\sin(\frac{x-\pi}{x+\pi}))=0$$

$$\therefore\lim_{x\rightarrow\pi}{\sin(\frac{x+\pi}{x-\pi})\sin(\frac{x-\pi}{x+\pi})}=0$$