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!
$$
\begin{align}
\lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})&=\lim_{x\to0^+}\frac{x^2\cos^2x-\sin^2x}{x^2\sin^2 x}\\
\vdots\\
&=\lim_{x\to0^+}\frac{-16\cos 2x+\ldots}{24\cos 2x+\ldots}\\
&=-\frac{2}{3}
\end{align}
$$

Without using the derivatives, or Taylor or o(little) or O(big),$$\lim \limits_{x \to 0}\left( \cot^2 x-{1 \over x^2}\right) = \lim \limits_{x \to 0} \frac{x^2\cos^2 x-\sin^2 x}{x^2\sin^2 x} = $$$$\lim \limits_{x \to 0} \frac{x\cos x+\sin x}{x}\cdot\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^3} $$$$\cdot\lim \limits_{x \to 0}\frac{x^2}{\sin^2 x}=\lim \limits_{x \to 0} \left( \cos x +\frac{\sin x}{x}\right)\cdot\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^3} $$$$= 2 \cdot \lim \limits_{x \to 0}\frac{\cos x-x\sin x- \cos x}{3x^2}=-\frac 23$$
