Help me with that problem, please.
$\lim_{x \to 0}\left ( \frac{1}{x^{2}}-\cot x\right )$
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 $\lim\limits_{x \to 0}x = 0$,
$\lim\limits_{x \to 0} = \left(\frac{1}{x^2} - \cot x\right) = \infty$
$\lim_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{\cos x}{\sin x}\right)=\infty,$
but
$\lim_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{\cos^{2}x}{\sin^{2}x}\right)=\frac{2}{3}.$
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. $