<s>[INST] Better approach to evaluate the limit $\lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})$ I solved it by rewriting the limit as indeterminate form $0/0$, then apply L'Hopital's rule 4 times, It was really lengthy and easy to make mistakes, If anyone got a better approach, please tell me!
$$
\begin{align}
\lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})&=\lim_{x\to0^+}\frac{x^2\cos^2x-\sin^2x}{x^2\sin^2 x}\\
\vdots\\
&=\lim_{x\to0^+}\frac{-16\cos 2x+\ldots}{24\cos 2x+\ldots}\\
&=-\frac{2}{3}
\end{align}
$$
 [/INST] Using $$\tan(x)=x+\frac{x^3}{3}+o(x^3)\quad \text{and}\quad \frac{1}{1+x}=1-x+o(x),$$ yields
\begin{align*}
\left(\cot(x)-\frac{1}{x}\right)\left(\cot(x)+\frac{1}{x}\right)&=\cot^2(x)-\frac{1}{x^2}\\
&=\frac{1}{\tan^2(x)}-\frac{1}{x^2}\\
&=\frac{1}{x^2+\frac{2}{3}x^4+o(x^4)}-\frac{1}{x^2}\\
&=\frac{1}{x^2}\left(\frac{1}{1+\frac{2}{3}x^2+o(x^2)}-1\right)\\
&=\frac{1}{x^2}\left(1-\frac{2}{3}x^2+1+o(x^2)\right)\\
&=-\frac{2}{3}+o(1).
\end{align*}
</s>