# limit of $\lim_{x \to 0}\left ( \frac{1}{x^{2}}-\cot x\right )$

Help me with that problem, please.

$$\lim_{x \to 0}\left ( \frac{1}{x^{2}}-\cot x\right )$$

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use that $\cot x = \frac{\cos x}{\sin x}$ and for the latter functions you know that $\sin x \sim x$ and $\cos x \sim 1-\frac12x^2$ at $x = 0$. And since this is homework, what have you tried? –  Ilya Mar 22 '12 at 16:28
I tried Lopital rule, but not sure that is right –  Unbelievable Me Mar 22 '12 at 16:36
Are you sure you don't mean $\cot^2x$? –  anon Mar 22 '12 at 16:49
@anon Yes I am sure –  Unbelievable Me Mar 22 '12 at 16:56
Then the limit does not exist. If it was $\cot^2x$ instead of $\cot x$ then the limit would be 2/3, and if it was $1/x$ instead of $1/x^2$ the limit would 0. –  anon Mar 22 '12 at 17:06

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

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 $\lim\limits_{x \to 0}x = 0$,

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

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$$\lim_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{\cos x}{\sin x}\right)=\infty,$$

but

$$\lim_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{\cos^{2}x}{\sin^{2}x}\right)=\frac{2}{3}.$$

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can you prove the second one ,please? –  Razvan Paraschiv Jul 29 at 0:33

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

$$\lim_{x\to 0}\left(\frac{1}{x^2} - \cot x \right) = \infty.$$

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