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

$$\lim\limits_{x \to 0} \left(\frac{1}{x^2} - \frac{1}{\tan x}\right) = \lim\limits_{x \to 0} -\left( \frac{x^4 \tan x - x^2 \tan^2 x}{x^4 \tan^2 x}\right) = 
-\lim\limits_{x \to 0} \frac{(x^2\tan x)(x^2-\tan x)}{(x^2 \tan x)(x^2 \tan x)}$$
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 $\lim\limits_{x \to 0}x = 0$,
$$\lim\limits_{x \to 0} = \left(\frac{1}{x^2} - \cot x\right) = \infty$$
