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.
 A: Hint: Our function is $\dfrac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$. The natural approach uses the Maclaurin series: We have $\sin x=x-\frac{x^3}{6}+O(x^5)$ and $\cos x=1-\frac{x^2}{2}+O(x^4)$. Now a little calculation does it. 
Remark: L'Hospital's Rule also works, but the calculation can be more unpleasant.  The fact that our denominator $x^2\sin^2 x$ behaves like $x^4$ near $0$ means that if we use L'Hospital's Rule mechanically, we will need $4$ applications of the Rule. So we use a couple of tricks.   
We start from the indeterminate form $\dfrac{1-x^2\cot^2 x}{x^2}$, and differentiate top and bottom. We want to find
$$\lim_{x\to 0}\frac{2x^2\cot x\csc^2 x-2x\cot^2 x}{2x}.$$
After making the obvious cancellation, and switching to sines and cosines, we find that we want
$$\lim_{x\to 0}\frac{x\cos x-\cos^2 x\sin x}{\sin^3 x}.$$
At this stage it helps to simplify, writing $\cos^2 x$ as $1-\sin^2 x$. So we want
$$\lim_{x\to 0}\frac{x\cos x-\sin x+\sin^3 x}{\sin^3 x}.$$
Split into the natural two parts, taking advantage of the cancellation $\frac{\sin^3 x}{\sin^3 x}=1$. Our limit is
$$1+\lim_{x\to 0} \frac{x\cos x-\sin x}{\sin^3 x}.$$
The rest is downhill. We have by L'Hospital's Rule,
$$\lim_{x\to 0} \frac{x\cos x-\sin x}{\sin^3 x}=\lim_{x\to 0} \frac{-x\sin x}{3\sin^2 x}.$$
We could apply L'Hospital's Rule twice more, but there is no point, the limit is $-\frac{1}{3}$, so our limit is $1-\frac{1}{3}$. 
A: Without hopital..
$$\cot(x)  \sim \frac1x - \frac{x}3$$
$$\cot(x)^2 \sim \frac1{x^2} +\frac{x^2}3 - \frac23$$
$$\begin{aligned}\lim_{x \to 0}\frac1{x^2}-\cot(x)^2&=\lim_{x \to 0} \frac1{x^2}-\frac1{x^2}-\frac{x^2}3+\frac23\\&=\lim_{x \to 0}\frac23 - \frac{x^2}3\\&= \frac23\end{aligned}$$
A: \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
$$
\lim_{x\rightarrow 0}\frac{\sin ^{2}x-x^{2}\cos ^{2}x}{x^{2}\sin ^{2}x}=%
\frac{\left( \frac{-1}{6}+\frac{1}{2}\right) \left( 1+1\right) }{\left(
1\right) ^{2}}=\frac{2}{3}.
$$
