Solving $ \lim_{x\to0}{\frac{1}{x^2}-\cot^2(x)}$ Evaluate:
$$ \lim_{x\to0}{\frac{1}{x^2}-\cot^2(x)}$$
My approach : 
$$\lim_{x\to0}{\frac{1}{x^2}-\frac{\cos^2(x)}{\sin^2(x)}}$$ 
$$ \lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} $$
Using $$\lim_{x\to0}\frac{\sin^2(x)}{x^2}=1 $$
$$ \lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^4} $$
$$ \lim_{x\to0}\frac{\sin^2(x)}{x^2}\cdot\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} $$
$$ \lim_{x\to0}\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} $$
Applying L Hopital,
$$\lim_{x\to0}\frac{2\sin(x)\cos(x)}{2x}=1$$
But the actual answer is $\frac{2}{3}$. What am I doing wrong here?

The following steps are not allowed
$$ \lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} \color{red}{=\lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^4} }= \lim_{x\to0}\frac{\sin^2(x)}{x^2}\cdot\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} =\\ \color{red}{= \lim_{x\to0}\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} }$$
since $\lim_{x\to0}\frac{\sin^2(x)}{x^2}=1 $ doesn't mean that $\frac{\sin^2(x)}{x^2}=1$.
We need to solve by l'Hopital or by Taylor's expansion starting from the first expression.
