<s>[INST] Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
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}
 [/INST] $$\lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right)=\lim_{x \to 0}\frac{\sin^2 x-x^2}{x^2\sin^2 x}=\lim_{x \to 0}\frac{(\sin x-x)(\sin x+x)}{x^4}$$
$$=\lim_{x \to 0}\frac{(\sin x+x)}{x}\lim_{x \to 0}\frac{x(\sin x-x)}{x^4}=\lim_{x \to 0}\frac{2x(\sin x-x)}{x^4}=\lim_{x \to 0}\frac{2(\sin x-x)}{x^3}$$
$$=\lim_{x \to 0}\frac{2(\cos x-1)}{3x^2}=\lim_{x \to 0}\frac{-2\sin x}{6x}=\frac{-1}{3}.$$
</s>