TITLE: Oblique asymptotes of $f(x)=\frac{x}{\arctan x}$
QUESTION [5 upvotes]: Find the oblique asymptotes of the function $\displaystyle f(x)=\frac{x}{\arctan x}$.


I tried finding the slope and $y$-intercept of the line going to $+\infty$ first, because the function is even (symmetrical in $x=0$). 
$\displaystyle a=\lim_{x\to\infty}\frac{f(x)}{x}=\lim_{x\to\infty}\frac{1}{\arctan x}=\frac{1}{\lim_{x\to\infty} \arctan x}=\frac{2}{\pi}$
Now, the slope is defined to be $\displaystyle b=\lim_{x\to\infty}f(x)-ax=\lim_{x\to\infty}\frac{x}{\arctan x}-\frac{2}{\pi}x$.
How do I evaluate this limit? I'm looking for just a hint. Also, if there is a better method for finding the asymptote, let me know.

REPLY [2 votes]: Let $y=mx+b$ be the oblique asymptote as $x\rightarrow\infty$.
Then $\displaystyle\lim_{x\to\infty}\left(\frac{x}{\arctan x}-mx-b\right)=0$, $\;\;$so $\displaystyle\lim_{x\to\infty}\left(\frac{x}{\arctan x}-mx\right)=b$  where
$\displaystyle\lim_{x\to\infty}\left(\frac{x}{\arctan x}-mx\right)=\lim_{x\to\infty}\left(\frac{1}{\arctan x}-m\right)x$ $\;\;$does not exist if $m\ne\frac{2}{\pi}$, since
$\displaystyle\lim_{x\to\infty}\left(\frac{1}{\arctan x}-m\right)=\frac{2}{\pi}-m$.$\;\;\;$
Therefore $\displaystyle m=\frac{2}{\pi}$, and $\displaystyle\lim_{x\to\infty}\left(\frac{x}{\arctan x}-\frac{2}{\pi}x\right)=$
$\displaystyle\lim_{x\to\infty}\frac{x\left(1-\frac{2}{\pi}\arctan x\right)}{\arctan x}=\lim_{x\to\infty}\frac{2}{\pi}\cdot{x\left(1-\frac{2}{\pi}\arctan x\right)}=\frac{2}{\pi}\lim_{x\to\infty}\frac{1-\frac{2}{\pi}\arctan x}{\frac{1}{x}}$
$=\displaystyle\frac{2}{\pi}\lim_{x\to\infty}\frac{-\frac{2}{\pi}\cdot\frac{1}{1+x^2}}{-\frac{1}{x^2}}=\frac{4}{\pi^{2}}\lim_{x\to\infty}\frac{x^2}{1+x^2}=\frac{4}{\pi^2},$ $\;\;$so $\displaystyle b=\frac{4}{\pi^2}$.
Thus $\displaystyle y=\frac{2}{\pi}x+\frac{4}{\pi^2}$ is one oblique asymptote, and the other is $\displaystyle y=-\frac{2}{\pi}x+\frac{4}{\pi^2}$ 
since the function is even.