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Find the oblique asymptotes of the function $\displaystyle f(x)=\frac{x}{\arctan x}$.

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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.

rae306
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3 Answers3

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For this specific case, let us change variable $x=\frac{1}{y}$. So $$\displaystyle \frac{x}{\arctan (x)}=\frac{1}{y \arctan(\frac{1}{y})}$$ and $y$ will go to $0$. Now, use $$\tan^{-1}(y)+\tan^{-1}(\frac{1}{y})=\frac{\pi}{2}$$. So the expression becomes $$\frac{1}{y (\frac{\pi}{2}-\tan^{-1}(y))}$$ Now, use Taylor epansion of $\tan^{-1}(y)$ built at $y=0$; it is $$\tan^{-1}(y)=y-\frac{y^3}{3}+\frac{y^5}{5}+O\left(y^6\right)$$ So, the denominator is $$\frac{\pi y}{2}-y^2+\frac{y^4}{3}-\frac{y^6}{5}+O\left(y^7\right)$$ Perform the long division for a few terms and get $$\frac{2}{\pi y}+\frac{4}{\pi ^2}+\frac{8 y}{\pi ^3}$$ Replace now $y$ by $\frac{1}{x}$ and obtain $$\frac{2 x}{\pi }+\frac{4}{\pi ^2}+\frac{8}{\pi ^3 x}$$ There are indeed faster ways to arrive to this result.

I hope this helps understanding the procedure (for this case).

Added later

If you think about it, the problem is very similar to the search of oblique asymptotes for $$\frac{a x^3+b x^2+c x+d}{e x^2+f x+g}$$ Performing long division, we should find $$\frac{b e-a f}{e^2}+\frac{a}{e}x +\cdots$$

Claude Leibovici
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  • Nice answer. Perhaps it would be easier for the OP to only use $\arctan{x} = x + o(x^2)$ (since I think it is sufficient) and to do another expansion for $\frac{1}{1-y}$ instead of "performing the long division". Anyway, +1. – Traklon Sep 19 '14 at 15:27
  • Very nice answer Claude! +1 ;-) Could you make a new answer where you also explain the faster way? I'm curious now :) – rae306 Sep 19 '14 at 15:29
  • The faster way is a direct expansion made at $x=\infty$. I prefered to give you this one because, I hope, you can do it in a quite systematic way. – Claude Leibovici Sep 19 '14 at 15:33
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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.

user84413
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1

From an intuitive perspective, I'd say that as $x$ gets very large $\arctan x \approx \pi /2$, so for large values of $x$ the function follows the line $y=(2/\pi)x$

Sheheryar Zaidi
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