<s>[INST] 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.
 [/INST] 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
$$
\begin{align}
\lim_{x\to0}\left(\cot^2(x)-\frac1{x^2}\right)
&=\lim_{x\to0}\left(\frac{\cos^2(x)}{\sin^2(x)}-\frac1{x^2}\right)\\
&=\lim_{x\to0}\frac{x^2\cos^2(x)-\sin^2(x)}{x^2\sin^2(x)}\\
&=\lim_{x\to0}\frac{x^2-x^2\sin^2(x)-\sin^2(x)}{x^2\sin^2(x)}\\
&=\lim_{x\to0}\frac{(x-\sin(x))(x+\sin(x))}{x^2\sin^2(x)}-1\\
&=\lim_{x\to0}\frac{x-\sin(x)}{x^3}\frac{x+\sin(x)}{x}\frac{x^2}{\sin^2(x)}-1\\
&=\underbrace{\lim_{x\to0}\frac{x-\sin(x)}{x^3}}_{\text{L'Hôpital}\,\times\,3}\underbrace{\left(1+\lim_{x\to0}\frac{\sin(x)}x\right)}_{\text{L'Hôpital}\,\times\,1}{\underbrace{\left(\lim\limits_{x\to0}\frac{\sin(x)}{x}\right)}_{\text{L'Hôpital}\,\times\,1}}^{-2}-1\\[3pt]
&=\frac16\cdot2\cdot1-1\\[3pt]
&=-\frac23
\end{align}
$$
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