Proving: $\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac1{x^2}\right)=\frac{\pi ^2}3$ without L'Hospital Evaluating
$$\lim_{x\to 0}\left(\frac{\pi ^2}{\sin ^2\pi x}-\frac{1}{x^2}\right)$$
with L'Hospital is so tedious. Does anyone know a way to evaluate the limit without using L'Hospital? I have no idea where to start.

Using Maclaurin expansion we get:
$$
\frac{\sin\pi x}{\pi x}
= \frac{\pi x - \frac16 (\pi x)^3 + O(x^5)}{\pi x}
= 1 - \frac{\pi^2 x^2}{6} + O(x^4)
\\
\frac{\pi x}{\sin\pi x} = 1 + \frac{\pi^2 x^2}{6} + O(x^4) 
\\
\left(\frac{\pi x}{\sin\pi x}\right)^2 = 1 + \frac{\pi^2 x^2}{3} + O(x^4) 
\\
\frac{\pi ^2}{\sin ^2\pi x}-\frac{1}{x^2}
= \frac{1}{x^2}\left[\left(\frac{\pi x}{\sin\pi x}\right)^2-1\right]
= \frac{\pi^2}{3} + O(x^2)
\to \frac{\pi^2}{3}
$$
