calculus - Evaluating $lim_{xrightarrowpi}frac{sin x}{x^2-pi ^2}$ without L'Hopital

I need to calculate the following limit (without using L'Hopital - I haven't gotten to derivatives yet):

$$\lim_{x\rightarrow\pi}\frac{\sin x}{x^2-\pi ^2}$$

We have $\sin$ function in the numerator so it looks like we should somehow make this similair to $\lim_{x\rightarrow 0} \frac{\sin x}{x}$. When choosing $t=x^2-\pi ^2$ we get $\lim_{t\rightarrow 0} \frac{\sin \sqrt{t+\pi ^2}}{t}$ so it's almost there and from there I don't know what to do. How to proceed further? Or maybe I'm doing it the wrong way and it can't be done that way?

Answer

Choosing the substitution $x - \pi = t$, with $t \to 0$, we have $$\lim_{x \to \pi}\frac{\sin x}{x^2 - \pi^2} = \lim_{t \to 0}\frac{\sin(t + \pi)}{t(t + 2\pi)} = \lim_{t \to 0}-\frac{\sin t}{t(t + 2\pi)} = -\frac1{2\pi}$$