$\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$ $\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$
I just can't do anything with this besides l'Hospital's rule (which doesn't seem to be a good idea). Can you help me, please?

Hint:
\begin{equation*}
\frac{\sin ^{2}x-x^{2}}{x^{2}\sin ^{2}x}=\left( \frac{\sin x-x}{x^{3}}%
\right) \left( \frac{\sin x+x}{x}\right) \left( \frac{x}{\sin x}\right) ^{2}
\end{equation*}
\begin{eqnarray*}
\lim_{x\rightarrow 0}\left( \frac{\sin x-x}{x^{3}}\right)  &=&-\frac{1}{6} \\
\lim_{x\rightarrow 0}\left( \frac{\sin x+x}{x}\right)  &=&1+1=2 \\
\lim_{x\rightarrow 0}\left( \frac{x}{\sin x}\right) ^{2} &=&1^{2}=1.
\end{eqnarray*}
