Simple approaches to prove that $\lim\limits_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)=-\frac13\ $? 
Find $\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$

My attempt:
$\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$ 
=$\lim_{x\to 0}(\frac{\sin^2 x -x^2}{x^2 \sin^2 x})$  ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{2\sin x \cos x -2x}{2x \sin^2 x+ 2x^2\sin x \cos x})$
=$\lim_{x\to 0}(\frac{\sin 2x -2}{2x \sin^2 x+ x^2\sin 2x})$  ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+2x\sin x \cos x)+ (2x\sin 2x+2x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x)+ (2x\sin 2x+2x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x)+ 2(x\sin 2x+x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{2\cos 2x}{2(\sin^2 x+x\sin 2x+x\sin 2x+x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{\cos 2x}{\sin^2 x+x\sin 2x+ x\sin 2x+x^2\cos 2x})$
=$\lim_{x\to 0}(\frac{\cos 2x}{\sin^2 x+2x\sin 2x+x^2\cos 2x})$  ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{-2\sin 2x}{2\sin x \cos x+2(\sin 2x+ 2x\cos 2x)+(2x\cos 2x-2x^2\sin 2x})$
=$\lim_{x\to 0}(\frac{-2\sin 2x}{3\sin 2x+6x\cos 2x-2x^2\sin 2x})$  ($\frac{0}{0}$ form)
Applying L'Hospital's Rule we get,
=$\lim_{x\to 0}(\frac{-4\cos 2x}{6\cos 2x+6(\cos 2x-2x\sin 2x)-2(2x\sin 2x+2x^2\cos 2x)})$
=$\lim_{x\to 0}(\frac{-4\cos 2x}{12\cos 2x-12x\sin 2x-4x\sin 2x-4x^2\cos 2x)})$
=$\frac{-4\cos 0}{12 \cos 0-0-0-0}$
=$\frac{-4}{12}$
=$\frac{-1}{3}$
My problem: This method is very lengthy and involves a lot of calculations. Is there is any other (better and efficient) method to evaluate this limit? 

$$\lim_{x\to 0}\frac{\sin x-x}{x^3}=\lim_{x\to 0}\frac{\cos x-1}{3x^2}=\lim_{x\to 0}\frac{-\sin x}{6x}=-\frac{1}{6}$$
\begin{align*}
\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)&=\lim_{x\to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}\\
&=\left(\lim_{x\to 0}\frac{\sin x-x}{x^3}\right)\left(\lim_{x\to 0}\frac{x\sin x+x^2}{\sin^2x}\right)\\
&=\left(\lim_{x\to 0}\frac{\sin x-x}{x^3}\right)\left(\lim_{x\to 0}\frac{x}{\sin x}+\lim_{x\to 0}\frac{x^2}{\sin^2x}\right)\\
&=\left(-\frac{1}{6}\right)(1+1)\\
&=-\frac{1}{3}
\end{align*}
