Evaluate $\lim_{x\to 0} \cot ^2 (x)-\frac{1}{x^2}$ 
Evaluate $\lim\limits_{x\to 0} \cot ^2 (x)-\dfrac{1}{x^2}.$ 

I was thinking of using L'Hôpital's Rule, but things got very, very ugly so I wasn't able to solve it. I know from experimentation that the limit is $-\dfrac{2}{3}$ though.
edit: i do not want to use a taylor expansion to solve this. That's too easy.
 A: Unacceptably Easy Answer (Before Question Edit)
$$
\begin{align}
\cot^2(x)
&=\frac{\cos^2(x)}{\sin^2(x)}\\
&=\frac{\left(1-\frac12x^2+O\!\left(x^4\right)\right)^2}{\left(x-\frac16x^3+O\!\left(x^5\right)\right)^2}\\
&=\frac1{x^2}\frac{1-x^2+O\!\left(x^4\right)}{1-\frac13x^2+O\!\left(x^4\right)}\\
&=\frac1{x^2}\left(1-\frac23x^2+O\!\left(x^4\right)\right)\\[6pt]
&=\frac1{x^2}\color{#C00}{-\frac23}+O\!\left(x^2\right)
\end{align}
$$

A More Difficult Approach Employing L'Hôpital
$$
\begin{align}
\lim_{x\to0}\left(\cot^2(x)-\frac1{x^2}\right)
&=\lim_{x\to0}\left(\frac{\cos^2(x)}{\sin^2(x)}-\frac1{x^2}\right)\\
&=\lim_{x\to0}\frac{x^2\cos^2(x)-\sin^2(x)}{x^2\sin^2(x)}\\
&=\lim_{x\to0}\frac{x^2-x^2\sin^2(x)-\sin^2(x)}{x^2\sin^2(x)}\\
&=\lim_{x\to0}\frac{(x-\sin(x))(x+\sin(x))}{x^2\sin^2(x)}-1\\
&=\lim_{x\to0}\frac{x-\sin(x)}{x^3}\frac{x+\sin(x)}{x}\frac{x^2}{\sin^2(x)}-1\\
&=\underbrace{\lim_{x\to0}\frac{x-\sin(x)}{x^3}}_{\text{L'Hôpital}\,\times\,3}\underbrace{\left(1+\lim_{x\to0}\frac{\sin(x)}x\right)}_{\text{L'Hôpital}\,\times\,1}{\underbrace{\left(\lim\limits_{x\to0}\frac{\sin(x)}{x}\right)}_{\text{L'Hôpital}\,\times\,1}}^{-2}-1\\[3pt]
&=\frac16\cdot2\cdot1-1\\[3pt]
&=-\frac23
\end{align}
$$
A: The L'Hôpital's rule can still be applied as shown below,
$$\lim\limits_{x\to 0} \left(\>\cot ^2 x-\dfrac{1}{x^2}\right)=\lim\limits_{x\to 0} \left(\frac1{\tan^2 x}-\dfrac{1}{x^2} \right)$$
$$=\lim\limits_{x\to 0}\frac{x-\tan x}{x^2\tan x}\cdot\left(\frac x{\tan x}+1\right) $$
$$=\lim\limits_{x\to 0}\frac{1-\sec^2 x}{2x\tan x+ x^2\sec^2x}\cdot2 
=\lim\limits_{x\to 0} \frac {-\frac{\tan x}x}{2+ \frac x{\tan x} \sec^2x}\cdot2 $$
$$=\frac {-1}{2+1\cdot 1}\cdot 2=-\frac23$$
where $\lim\limits_{x\to 0}\frac{\tan x}x =1$ is used in above evaluation.
A: We have that
$$\tan\left(x\right)=x+\frac1{3}x^3+o\left(x^3\right)$$
and
$$\cot x =\frac1{\tan x}=\frac1x\left(1+\frac1{3}x^2+o\left(x^2\right)\right)^{-1}=\frac1x-\frac13x-o(x)$$
$$\cot^2 x =\left(\frac1x-\frac13x+o(x)\right)^2=\frac1{x^2}-\frac23+o(1)$$
and then
$$\cot^2 x-\frac1{x^2}=-\frac 23 +o(1)\to -\frac23$$
A: Since $\sin x=x(1-\frac16x^2+o(x^2))$ and $\cos x=1-\frac12x^2+o(x^2)$,$$\cot^2x=\cos^2x\sin^{-2}x=\frac{1}{x^2}\left(1+2\left(\frac16-\frac12\right)x^2+o(x^2)\right)=\frac{1}{x^2}-\frac23+o(1).$$
