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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?2012-03-22
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    I tried Lopital rule, but not sure that is right2012-03-22
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    Are you sure you don't mean $\cot^2x$?2012-03-22
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    @anon Yes I am sure2012-03-22
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    Then the limit [does not exist](http://www4c.wolframalpha.com/Calculate/MSP/MSP54361a0h0i0e1b0a315d00003b22h085b78f8ib6?MSPStoreType=image/gif&s=30&w=300&h=192&cdf=RangeControl). 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 0.2012-03-22
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    But, you could say the limit has value $\infty$. If infinite limits are not allowed, the problem is trivial (consider the limit from the left). L'Hopital's rule does work nicely here.2012-03-22
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    @ David Mitra Thanks a lot, but how can I to prove that limit value it's infinite?2012-03-22
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    Did you try using Ilya's hint? What can you say about $\frac{1}{x^2} - \frac{1}{x}$?2012-03-22
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    Or, write ${1\over x^2}-{\cos x\over \sin x}={\sin x -x^2\cos x\over x^2\sin x}$ and use L'Hopital. Ilya's suggestion would be quicker, though. Be careful with signs...2012-03-22
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    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

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

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    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

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