<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] $\cot^2x-\frac1{x^2}=\csc^2x-1-\frac1{x^2}=\frac1{\sin^2x}-\frac1{x^2}-1=\frac{x^2-\sin^2x}{x^2\sin^2x}-1=\frac{(x-\sin x)(x+\sin x)}{x^4\frac{\sin^2x}{x^2}}-1$
In the numerator, using $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}...$, thus,
$\frac{(\frac{x^3}{3!}-\frac{x^5}{5!}+...)(2x-\frac{x^3}{3!}+...)}{x^4\frac{\sin^2x}{x^2}}-1$
Using $\lim_{x\to0}\frac{\sin x}{x}=1$, we get,
$$\lim_{x\to0}\frac{(\frac{x^3}{3!}-\frac{x^5}{5!}+...)(2x-\frac{x^3}{3!}+...)}{x^4}-1\\=\lim_{x\to0}\frac{(\frac1{3!}-\frac{x^2}{5!}+..)(2-\frac{x^2}{2!}+...)}1-1\\=\frac26-1=-\frac23$$
</s>