limit of $\lim_{x \to 0}\left ( \frac{1}{x^{2}}-\cot x\right )$ Help me with that problem, please.
$$\lim_{x \to 0}\left ( \frac{1}{x^{2}}-\cot x\right )$$

Note that for $x>0$ near $0$ we have
$$
\frac{1}{x^2} - \cot x  = \frac{1}{x^2}-\frac{\cos x}{\sin x} \geq \frac{1}{x^2}-\frac{1}{\sin x} = \frac{\sin x - x^2}{x^2 \sin x}.
$$
Then we have
$$
\lim_{x\to 0^+}\frac{\sin x - x^2}{x^2 \sin x} 
\ \operatorname*{=}^{\small\mathrm{L'H}}\ \lim_{x\to 0^+} \frac{\cos x-2x}{2x\sin x + x^2\cos x } = \infty 
$$
so it follows by the squeeze theorem that
$$
\lim_{x\to 0^+}\left(\frac{1}{x^2} - \cot x \right) = \infty.
$$
For $x<0$ near $0$ we have $\cot x < 0$ so $$\frac{1}{x^2}-\cot x \geq \frac{1}{x^2} \to \infty$$
and so
$$
\lim_{x\to 0}\left(\frac{1}{x^2} - \cot x \right) = \infty.
$$
