<s>[INST] Indeterminate limit that is supposed to be solved with De L'Hospital's rule Last week my Maths teacher gave the class this exercise taken from our text book. We are working on De L'Hospital's rule at the moment and this exercise is from that part of the book so everybody assumed that was the right procedure to solve it. One week later and nobody has been able to get to the right solution (even using other procedures).
This is the exercise:

Solve
  $$
\lim_{x \to 0} \frac{1}{x^{2}} - \cot^{2}{x}
$$

According to the book and the mighty WolframAlpha the solution is $\frac{2}{3}$ but I can't get anywhere near it. The only solution I were able to get was a $-\infty$, which I got by transforming the $\cot^{2} x $ in $\frac{\cos^{2} x}{\sin^{2} x}$ and using De L'Hospital's.
EDIT: I'm asking just out of curiosity, I don't have to turn in this exercise and I'm not asking to avoid doing my homework myself.
 [/INST] \begin{eqnarray*}
\frac{\sin ^{2}x-x^{2}\cos ^{2}x}{x^{2}\sin ^{2}x} &=&\frac{\left( \sin
x-x\cos x\right) \left( \sin x+x\cos x\right) }{x^{2}\sin ^{2}x} \\
&=&\frac{\frac{\left( \sin x-x+x-x\cos x\right) }{x^{3}}\frac{\left( \sin
x+x\cos x\right) }{x}}{\frac{x^{2}\sin ^{2}x}{x^{4}}} \\
&=&\frac{\left( \frac{\sin x-x}{x^{3}}+\frac{1-\cos x}{x^{2}}\right) \left( 
\frac{\sin x}{x}+\cos x\right) }{\left( \frac{\sin x}{x}\right) ^{2}}
\end{eqnarray*}
$\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}}\overset{HRule}{=}%
\lim_{x\rightarrow 0}\frac{\cos x-1}{3x^{2}}\overset{HRule}{=}%
\lim_{x\rightarrow 0}\frac{-\sin x}{6x}\overset{HRule}{=}\lim_{x\rightarrow
0}\frac{-\cos x}{6}=\frac{-1}{6}.$
$\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}\overset{HRule}{=}%
\lim_{x\rightarrow 0}\frac{\sin x}{2x}\overset{HRule}{=}\lim_{x\rightarrow 0}%
\frac{\cos x}{2}=\frac{1}{2}$
$\lim_{x\rightarrow 0}\frac{\sin x}{x}\overset{HRule}{=}\lim_{x\rightarrow 0}%
\frac{\cos x}{1}=1.$ It follows that
$$
\lim_{x\rightarrow 0}\frac{\sin ^{2}x-x^{2}\cos ^{2}x}{x^{2}\sin ^{2}x}=%
\frac{\left( \frac{-1}{6}+\frac{1}{2}\right) \left( 1+1\right) }{\left(
1\right) ^{2}}=\frac{2}{3}.
$$
</s>