limit of $\lim_{x \to 0}\left ( \frac{1}{x^{2}}-\cot x\right )$ Help me with that problem, please.
$$\lim_{x \to 0}\left ( \frac{1}{x^{2}}-\cot x\right )$$
 A: $$\lim\limits_{x \to 0} \left(\frac{1}{x^2} - \frac{1}{\tan x}\right) = \lim\limits_{x \to 0} -\left( \frac{x^4 \tan x - x^2 \tan^2 x}{x^4 \tan^2 x}\right) = 
-\lim\limits_{x \to 0} \frac{(x^2\tan x)(x^2-\tan x)}{(x^2 \tan x)(x^2 \tan x)}$$
Cancelling out terms:
$$-\lim\limits_{x \to 0} \frac{x^2 - \tan x}{x^2 \tan x}$$
Apply L'Hopitals Rule
$$-\lim\limits_{x \to 0}\frac{x \cos2x + x - 1}{x(x+\sin 2x)} =-\frac{\lim\limits_{x \to 0}x + \lim\limits_{x \to 0}x \cos 2x - 1}{\lim\limits_{x \to 0}x(x+\sin 2x)} =-\frac{-1}{\lim\limits_{x \to 0}x(x+\sin2x)}$$ 
The limit of the products is the product of the limits.
$$\frac{1}{\lim\limits_{x \to 0}x(x+\sin2x)} = \frac{1}{(\lim\limits_{x \to 0}x)(\lim\limits_{x \to 0}(x + \sin 2x))}$$
Since $\lim\limits_{x \to 0}x = 0$,
$$\lim\limits_{x \to 0} = \left(\frac{1}{x^2} - \cot x\right) = \infty$$
A: $$\lim_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{\cos x}{\sin x}\right)=\infty,$$
but
$$\lim_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{\cos^{2}x}{\sin^{2}x}\right)=\frac{2}{3}.$$
A: 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
$$
\lim_{x\to 0}\left(\frac{1}{x^2} - \cot x \right) = \infty.
$$
