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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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    @AndreasHagen Sorry, that's my stupid misunderstanding. I'm too used to thinking of asymptotes as horizontal or vertical.2012-10-26

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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.