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 right – 2012-03-22
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Are you sure you don't mean $\cot^2x$? – 2012-03-22
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@anon Yes I am sure – 2012-03-22
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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