<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] As noticed in the comments, we are allowed to proceed as follows
$$\lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right)= \lim_{x \to 0} \left( \frac{\sin^2 x-x^2}{x^2\sin^2 x} \right)=\lim_{x \to 0} \left( \frac{\sin x+x}{x\sin x} \right)\left( \frac{\sin x-x}{x\sin x} \right)=\ldots$$
but we are not allowed to proceed as follows
$$\ldots=\lim_{x \to 0} \left( \frac{\sin x+x}{x\sin x} \right)\lim_{x \to 0}\left( \frac{\sin x-x}{x\sin x} \right)=\ldots$$
when one or both limits do not exist or the product leads to an undefined expression.
Notably in that case by l'Hopital we obtain 
$$\ldots=\lim_{x \to 0} \frac {\cos x+1}  {\sin x+x\cos x}\cdot \lim_{x \to 0} \frac {\cos x-1}  {\sin x+x\cos x}=\ldots$$
and the LHS limit, in the form $\frac 2 0$, doesn't exist while the RHS limit is equal to zero. 
Therefore the initial step in that case doesn't work.
Note that in any case also the following step
$$  \ldots=\lim_{x \to 0} (\cos x+1)\,\lim_{x \to 0} \frac {\cos x-1}  {(\sin x+x\cos x)^2}=\ldots$$
is not allowed since once we have divided the original limit as the product of two distinct limits we need to operate separetely on each of them when using l'Hopital or Taylor's series. Only when we have calculated the limit for each expression we know whether the initial step was allowed or not.
See also the related Analyzing limits problem Calculus (tell me where I'm wrong).

In that case, following for example the hint given by mrs, a correct way to proceed by l'Hopital is as follows
$$\lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right)
= \lim_{x \to 0}\left(\frac{\sin^2 x-x^2}{x^4}\cdot\frac{x^2}{\sin^2 x}\right)
\stackrel{?} = \lim_{x \to 0}\frac{\sin^2 x-x^2}{x^4}\cdot\lim_{x \to 0}\frac{x^2}{\sin^2 x }=\ldots$$
and since, using l'Hopital for each part, we have
$$\lim_{x \to 0}\frac{\sin^2 x-x^2}{x^4}=\lim_{x \to 0}\frac{\sin 2x-2x}{4x^3}=\lim_{x \to 0}\frac{2\cos 2x-2}{12x^2}=\lim_{x \to 0}\frac{-4\sin 2x}{24x}=\lim_{x \to 0}\frac{-8\cos 2x}{24}=-\frac13$$
$$\lim_{x \to 0}\frac{x^2}{\sin^2 x }=\lim_{x \to 0}\frac{2x}{\sin 2x }=\lim_{x \to 0}\frac{2}{2\cos 2x }=1$$
we see that the initial step is allowed and then we can conclude that
$$\ldots= \lim_{x \to 0}\frac{\sin^2 x-x^2}{x^4}\cdot\lim_{x \to 0}\frac{x^2}{\sin^2 x }=-\frac13\cdot 1 =-\frac13$$
Note finally that some intermediate step can be highly simplified using the standard limit $\lim_{x \to 0}\frac{\sin x }x=1$.
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