TITLE: Calculate the limit $\lim_{x\to 0} \left(\frac 1{x^2}-\cot^2x\right)$
QUESTION [8 upvotes]: The answer of the given limit is $2/3$, but I cannot reach it. I have tried to use the L'Hospital rule, but I couldn't drive it to the end. Please give a detailed solution!
$$\lim_{x\to 0}  \left(\dfrac 1{x^2}-\cot^2x\right)$$

REPLY [5 votes]: We can proceed in the following manner
\begin{align}
L &= \lim_{x \to 0}\left(\frac{1}{x^{2}} - \cot^{2}x\right)\notag\\
&= \lim_{x \to 0}\left(\frac{1}{x^{2}} - \frac{\cos^{2}x}{\sin^{2}x}\right)\notag\\
&= \lim_{x \to 0}\frac{\sin^{2}x - x^{2}\cos^{2}x}{x^{2}\sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\sin^{2}x - x^{2}\cos^{2}x}{x^{4}}\cdot\frac{x^{2}}{\sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{\sin^{2}x - x^{2}\cos^{2}x}{x^{4}}\cdot 1^{2}\notag\\
&= \lim_{x \to 0}\frac{\sin x - x\cos x}{x^{3}}\cdot\frac{\sin x + x\cos x}{x}\notag\\
&= \lim_{x \to 0}\frac{\sin x - x\cos x}{x^{3}}\cdot\left(\frac{\sin x}{x} + \cos x\right)\notag\\
&= 2\lim_{x \to 0}\frac{\sin x - x\cos x}{x^{3}}\notag\\
&= 2\lim_{x \to 0}\frac{\cos x - \cos x + x\sin x}{3x^{2}}\text{ (via LHR)}\notag\\
&= \frac{2}{3}\lim_{x \to 0}\frac{\sin x}{x}\notag\\
&= \frac{2}{3}\notag
\end{align}
As I have told often on MSE, LHR is best used with a combination of standard limits and rules of "algebra of limits". When you have to differentiate too many times in applying LHR its a signal that you are going on a laborious path and perhaps there is a simpler route. The above solution looks a bit long because of detailed steps otherwise its very simple.