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.

Yet another approach: as it's an even function, assume $x>0$. Cut a sector in a radius-$\sqrt{2}$ angle subtending $\pi x$ radians at the centre so $\pi x-\sin\pi x$ is the area in the sector outside the triangle with the same vertices. We'll approximate the arc as a parabola, in Cartesian coordinates with the line segment of the same endpoints a part of the $X$-axis, with extrema at $X=\pm\sqrt{2}\sin\frac{\pi x}{2}\sim\pm\frac{\pi x}{\sqrt{2}}$. The peak is at$$X=0,\,Y=\sqrt{2}(1-\cos\frac{\pi x}{2})=2\sqrt{2}\sin^2\frac{\pi x}{4}\sim\frac{\pi^2x^2\sqrt{2}}{8}.$$At leading order, the parabola is $Y=\frac{\sqrt{2}}{4}(\pi^2x^2/2-X^2)$, so the area below it is$$\int_{-\pi x/\sqrt{2}}^{\pi x/\sqrt{2}}\frac{\sqrt{2}}{4}(\pi^2x^2/2-X^2)dX=\frac{\pi^3x^3}{6}.$$So for small $x$,$$\pi x-\sin\pi x\sim\frac{\pi^3x^3}{6}\implies\frac{1}{\sin\pi x}-\frac{1}{\pi x}\sim\frac{\pi^2x^2}{6\sin\pi x}\sim\frac{\pi x}{6}\implies\frac{1}{\sin^2\pi x}-\frac{1}{\pi^2x^2}\sim\frac{\pi x}{2}\cdot\frac{2}{\pi x}=\frac13.$$
