$$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\begin{align}
\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)&=\lim_{x\to0}\frac{(\sin x+x)(\sin x-x)}{(x\sin x)(x\sin x)} \\[1ex]
&=\lim_{x\to0}\left(\frac{\sin x+x}{x\sin x}\right)\lim_{x\to0}\left(\frac{\sin x-x}{x\sin x}\right) \\[1ex]
&=\lim_{x\to0}\left(\frac{\cos x+1}{\sin x+x\cos x}\right)\lim_{x\to0}\left(\frac{\cos x-1}{\sin x+x\cos x}\right) \\[1ex]
&=\lim_{x\to0}\:(\cos x+1)\,\lim_{x\to0}\left(\frac{\cos x-1}{(\sin x+x\cos x)^2}\right) \\[1ex]
&=\lim_{x\to0}\frac{-\sin x}{(\sin x+x\cos x)(2\cos x-x\sin x)} \\[1ex]
&=-\lim_{x\to0}\left[\frac{1}{1+\cos x\left(\frac{x}{\sin x}\right)}\right]\left(\frac{1}{2\cos x-x\sin x}\right) \\[1ex]
&=-\frac{1}{2}\left[\lim_{x\to0}\,\frac{1}{1+\cos x}\right] \\[1ex]
&=-\frac{1}{4}
\end{align}
Best Answer
By l'Hopital we have
$$\lim_{x \to 0}\frac{1}{x^2} - \frac{1} {\sin^2 x} =\lim_{x \to 0}\frac{\sin^2 x-x^2}{x^2\sin^2 x}$$
$$\stackrel{H.R.}=\lim_{x \to 0}\frac{\sin 2x-2x}{2x\sin^2 x+x^2\sin 2x }$$
$$\stackrel{H.R.}=\lim_{x \to 0}\frac{2\cos 2x-2}{2\sin^2 x+2x\sin 2x+2x\sin 2x +2x^2\cos 2x}$$
$$\stackrel{H.R.}=\lim_{x \to 0}\frac{-4\sin 2x}{2\sin 2 x+8x\cos 2x+4 \sin 2x+4x\cos 2x-4x^2\sin 2x}$$
$$\stackrel{H.R.}=\lim_{x \to 0}\frac{-8\cos 2x}{12\cos 2 x+8\cos 2x-16x \sin 2x-8x\sin 2x+4\cos 2x-8x\sin 2x-8x^2\cos2x}$$
$$=\lim_{x \to 0}\frac{-8\cos 2x}{24\cos 2 x-32x \sin 2x-8x^2\cos2x} =\frac{-8}{24-0-0}=-\frac13$$