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I'm struggling with this limit :

$$ \lim_{x\to 0} {(x \cot x)-1 \over x^2} $$

It's in the $0\over0$ form, so I tried to use L'Hospital's rule: $$ \lim_{x\to 0} {((x \cot x)-1)' \over (x^2)'} = \lim_{x\to 0} { \cot x - x \csc^2 x \over 2x} = {\frac 10 - \frac 10 \over 0} $$ I'm not sure how to continue. I tried to derivate more, but only got similar expressions or $\frac 00$. The result should be $-\frac13 $, so guess I'm doing something wrong. I would appreciate if somebody could tell me how to solve this one.

Thanks for help.

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    $\begingroup$ Have you learned about Taylor series? $\endgroup$ – DanZimm Jan 2 '15 at 18:33
  • $\begingroup$ No, but ill look it up. Thanks for the tip. $\endgroup$ – user204416 Jan 2 '15 at 18:40
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$$\frac{x\cot x-1}{x^2}=\frac{x\cos x-\sin x}{x^3}\cdot\frac{x}{\sin x}=\left(\frac{x-\sin x}{x^3}+\frac{\cos x-1}{x^2}\right)\cdot\frac{x}{\sin x}$$ hence the limit of the RHS as $x\to 0$ is $\frac{1}{6}-\frac{1}{2}=\color{red}{-\frac{1}{3}}$.

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As stated in my comment Taylor series are a pretty good idea here. We have $$ \cot x = x^{-1} - \frac{1}{3} x + o(x^3) $$ You should be able to take it from here noting that $o(x^3) / x^2 \to 0$ as $x \to 0$.

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