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?
 A: Although $\frac{\sin^2x}{x^2}\to1$, $\frac{\sin^2x}{x^2}\frac{1}{x^2}-\frac{\cos^2x}{x^2}$ isn't asymptotic to $\frac{1}{x^2}-\frac{\cos^2x}{x^2}$, because $$\lim_{x\to0}\left(\frac{\sin^2x}{x^2}\frac{1}{x^2}-\frac{1}{x^2}\right)=\lim_{x\to0}\frac{\frac{\sin^2x}{x^2}-1}{x^2}=\lim_{x\to0}\frac{\left(1-\frac16x^2\right)^2-1}{x^2}=-\frac13.$$The technique I'd advise is $$\cot x=\frac1x\frac{1-x^2/2+o(x^2)}{1-x^2/6+o(x^2)}=\frac1x\left(1-\frac13x^2+o(x^2)\right)\\\implies\frac{1}{x^2}-\cot^2x=\frac{1}{x^2}\left(\frac23x^2+o(x^2)\right)\to\frac23.$$
A: 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.
A: $$L=\lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\cot^2[x] \right)= \frac{1}{x^2}-\frac{1}{(x+x^3/3+O(x^5))^2}$$ Let us use $(1+z)^{\nu} \approx 1+\nu z, ~if ~$z$ <<1.$
$$\Rightarrow L=\frac{1}{x^2}-\frac{1}{x^2} (1+x^2/3+O(x^4)^{-2}=\frac{1}{x^2}-\frac{1}{x^2} (1-2x^2/3+O(x^4))= \lim_{x \rightarrow 0} \frac{2}{3}+O(x^2)=\frac{2}{3}. $$
