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I'm supposed to find asymptotes for $x\cdot\arctan(x)$ for a homework assignment.

Are there any theorems regarding this I can utilize to find the equation for the asymptotes, or is this one of those solve case by case things, where this particular equation can be solved in some neat way, but there is no easy universal way to find asymptotes like this?

I know it should be:

$$\pm \frac{\pi}{2} x - 1$$

but "how do I show this?" is essentially my question.

Thanks in advance

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Do you mean $1/\tan(x)$ or $\arctan(x)$ multiplied by $x$? – m0nhawk Oct 26 '12 at 11:24
arctan, didn't think about the confusion it might lead to not specify :) – Andreas Hagen Oct 26 '12 at 11:27
$x \cdot \arctan(x)$ doesn't have any asymptotes. – SiliconCelery Oct 26 '12 at 11:47
Well yes it does. Look here: Same goes for the other side as well, even though WA does not show it. – Andreas Hagen Oct 26 '12 at 11:53
@AndreasHagen Sorry, that's my stupid misunderstanding. I'm too used to thinking of asymptotes as horizontal or vertical. – SiliconCelery Oct 26 '12 at 12:34
up vote 1 down vote accepted

I am sure the following is in your textbook and/or has been explained in class.

Let $f\colon\mathbb{R}\to\mathbb{R}$ be a function.

  1. If $\lim_{x\to+\infty}f(x)=a$, then $y=a$ is a horizontal asymptote (similarly for $x\to-\infty$.)
  2. If for some $b\in\mathbb{R}$ $\lim_{x\to b^+}f(x)=\pm\infty$, then $x=b$ is a vertical asymptote (similarly for $x\to b^-$.)
  3. If $\lim_{x\to+\infty}\frac{f(x)}{x}=m$, $m\ne0$, and $\lim_{x\to+\infty}(f(x)-m\,x)=c$, then $y=m\,x+c$ is an oblique asymptote similarly for $x\to-\infty$.)

Apply it to $x\arctan x$. The only difficulty is finding $\lim_{x\to-\infty}x\,(\arctan x-\frac\pi2)$, which can be done using L'Hôpital's rule.

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