Calculate the limit $\lim_{x\to 0} \left(\frac 1{x^2}-\cot^2x\right)$ 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!
$$\lim_{x\to 0}  \left(\dfrac 1{x^2}-\cot^2x\right)$$

Here's a non-series solution.
$$\lim_{x \rightarrow 0} \left(\frac{1}{x^2} - \cot^2(x)\right) = \lim_{x \rightarrow 0} \frac{\sin^2(x) - x^2\cos^2(x)}{x^2\sin^2(x)}$$
Instead of directly doing l'Hospital, we can make our life easier using $\lim_{x \rightarrow 0}\frac{\sin(x)}{x} = 1$ Or rather, $\lim_{x \rightarrow 0}\frac{x}{\sin(x)} = 1$:
$$\lim_{x \rightarrow 0} \frac{\sin^2(x) - x^2\cos^2(x)}{x^2\sin^2(x)} = \lim_{x \rightarrow 0}\left(\frac{x}{\sin(x)}\right)^2\left(\frac{\sin^2(x) - x^2\cos^2(x)}{x^4}\right) = \lim_{x \rightarrow 0} \frac{\sin^2(x) - x^2\cos^2(x)}{x^4}$$
Now l'Hospital. Note that we can freely factor out constants, and also $\cos(x)$ (as it approaches 1 as $x \rightarrow 0$):
\begin{align*} \lim_{x \rightarrow 0} \frac{\sin^2(x) - x^2\cos^2(x)}{x^4} &= \lim_{x \rightarrow 0} \frac{2\sin(x)\cos(x) - 2x\cos^2(x) + 2x^2\sin(x)\cos(x)}{4x^3} \\
&= \frac{1}{2}\lim_{x \rightarrow 0}\frac{\sin(x) - x\cos(x) + x^2\sin(x)}{x^3} \\
&= \frac{1}{2}\lim_{x \rightarrow 0}\frac{\cos(x) - \cos(x) + x\sin(x) + 2x\sin(x) + x^2\cos(x)}{3x^2} \\
&= \frac{1}{6}\lim_{x \rightarrow 0}\frac{3x\sin(x) + x^2\cos(x)}{x^2} \\
&= \frac{1}{6}\left(\lim_{x \rightarrow 0}3\frac{\sin(x)}{x} + \lim_{x \rightarrow 0} \cos(x)\right) \\
&= \frac{1}{6}(3 + 1) = \frac{2}{3}\end{align*}
