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Help me with that problem, please.

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

  • 0
    Just at a glance, the two-sided limit either does not exist in any sense or is positive $\infty$ if that sense is allowed. This is because $\frac{1}{x^2}\to\infty$ as $x\to0^-$ while $\cot(x)\to-\infty$ as $x\to0^-$.2012-07-09

3 Answers 3

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

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

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

0

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

but

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

  • 0
    can you prove the second one ,please?2015-07-29
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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. $