How to find trigonometric limit $ \lim_{x\to π} \frac{\sin x}{x^2 - π^2} $

Question

how to solve this problem? (without using l'Hopital rule) $$ \lim_{x\to π} \frac{\sin x}{x^2 - π^2} $$ I have no idea how to transform that so that I have no expression $\frac{0}{0}$
I suspect that calculating limit $\lim_{x\to π^{-}}$ or $\lim_{x\to π^+}$ does not works there.
Thanks for helping

Answer 1

Rewrite as $$ \lim_{x\to \pi}\frac{\sin x}{x^2-\pi^2}=\lim_{x\to \pi}\frac{\sin x-\sin \pi}{x-\pi}\times \lim_{x\to \pi}\frac{1}{x+\pi} $$ The first limit is (by definition) the derivative of $\sin $ at $x=\pi$. The second limit is easy to compute.

Answer 2

Hint: $$\lim_{x\to\pi}\frac{\sin x}{x^2-\pi^2}=\left(\lim_{x\to\pi}\frac1{x+\pi}\right)\left(\lim_{x\to\pi}\frac{\sin x}{x-\pi}\right).$$