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

\begin{align}
\frac{\sin(x)}{x^2 - \pi^2} = \frac{x}{x^2(1-\pi^2/x^2)} \prod_{n=1}^{\infty}\left(1 -\frac{x^2}{n^2\pi^2}\right) = \frac{1}{x} \frac{1-x^2/\pi^2}{1-\pi^2/x^2} \prod_{n=2}^{\infty} \left(1 -\frac{x^2}{n^2\pi^2}\right)
\end{align}
Since no L'Hospital available, we consider of the sign of numerator and denominator as $x \to \pi^{+}$ and $x \to \pi^-$ to obtain
\begin{align}
\lim_{x\to \pi}\frac{1-x^2/\pi^2}{1-\pi^2/x^2}  = -1
\end{align}
and hence
\begin{align}
\lim_{x\to \pi }\frac{\sin(x)}{x^2 - \pi^2}
=
-\frac{1}{\pi} \prod_{n=2}^{\infty} \left(1 -\frac{1}{n^2}\right) 
= -\frac{1}{2\pi}
\end{align}
