Finding the oblique asymptote of $f(x)=2x\tan^{-1}(x)$

Question

I have problems with the computation of the oblique asymptote for the function

$$f(x)=2x\tan^{-1}(x)$$

I started by stating that $y = mx + b$ is the equation of the oblique asymptote. Next I looked at the following limit

$$\lim_{x \rightarrow \infty } \left(2x \tan^{-1}(x) - mx - b\right) = 0 $$

from where I know that since $m$ and $b$ are constants, that I must have:

$$\lim_{x \rightarrow \infty }\left(2x\tan^{-1}(x) - mx\right) = b$$ thus we must have $$\lim_{x \rightarrow \infty }\left(2x\tan^{-1}(x) - mx\right) = 0$$ and hence $$x\lim_{x \rightarrow \infty }\left(2\tan^{-1}(x) - m\right) = 0$$ which does not exists unless we have that $m=\pi$. From here I do not know how to proceed. I tried plugging in my value for $m$ - but without luck.

I found a similar question here, Oblique asymptotes of $f(x)=\frac{x}{\arctan x}$ , and tried to follow it, but without luck.

Extra:

Best regards

Answer 1

Let $y=mx+b$ be the oblique asymptote as $x\rightarrow\infty$.

Then $\displaystyle\lim_{x\to\infty}\left(2x\tan^{-1}(x)-mx-b\right)=0$,

so $\displaystyle\lim_{x\to\infty}\left(2x\tan^{-1}(x)-mx\right)=b$

We have that $\displaystyle\lim_{x\to\infty}\left(2x\tan^{-1}(x)-mx\right)=\lim_{x\to\infty}\left(2\tan^{-1}(x)-m\right)x$ doesn't exist if $\displaystyle{\lim_{x\to\infty}\left(2\tan^{-1}(x)-m\right)\neq 0 \Rightarrow \lim_{x\to\infty}\tan^{-1}(x)\neq \frac{m}{2}\Rightarrow \frac{\pi}{2}\neq \frac{m}{2} \Rightarrow m\neq \pi}$.

So, $\displaystyle m=\pi$, and then $\displaystyle\lim_{x\to\infty}\left(2x\tan^{-1}(x)-\pi x\right)=\lim_{x\rightarrow 0}\left(2\frac{1}{x}\tan^{-1}\left (\frac{1}{x}\right )-\pi \frac{1}{x}\right )=\lim_{x\rightarrow 0}\left(\frac{2\tan^{-1}\left (\frac{1}{x}\right )-\pi }{x}\right ) \ \overset{\text{ De l'Hospital }}{ = } \ \lim_{x\rightarrow 0}\left(2\frac{-1}{x^2+1}\right )=-2$

so $b=-2$.

Therefore, $\displaystyle y=\pi x-2$ is the oblique asymptote.

Answer 2

$ax+b$ being an asymptote to $2x\tan^{-1}x$ is equivalent to the statement;

$$2x\tan^{-1}x-ax-b=o(1)$$

as $x\to\infty$. Now, dividing by $x$ tells us

$$2\tan^{-1}x-a=o(1)\implies a=\lim_{x\to\infty}2\tan^{-1}x=\pi$$

This leads us to a refinement of our first statement, seeing that

$$2x\tan^{-1}x-\pi x-b=o(1)\implies b=\lim_{x\to\infty}2x\tan^{-1}x-\pi x$$

We now split this expression as $2x\left(\tan^{-1}x-\frac{\pi}{2}\right)$, and note that for $x>0$, $\tan^{-1}x=\frac{\pi}{2}-\tan^{-1}\left(\frac{1}{x}\right)$, whence:

$$b=\lim_{x\to\infty}\left(-2x\tan^{-1}\left(\frac{1}{x}\right)\right)$$

Let $x=\frac{1}{y}$ to reframe this limit as $\lim_{y\to0}\frac{-2\tan^{-1}y}{y}$ and use L'Hopital's Rule / the definition of the derivative / any of a number of other methods to retrieve this limit as $-2$.

We are thus assured that:

$$2x\tan^{-1}x=\pi x-2+o(1)$$

i.e. that the asymptote as $x\to\infty$ of $2x\tan^{-1}x$ is $\pi x-2$.

Answer 3

Use an asymptotic expansion.

First set $x=\dfrac1u$ and remember $\;\arctan x=\begin{cases}\dfrac\pi2-\arctan u &\text{if}\enspace x>0,\\ -\dfrac\pi2-\arctan u &\text{if}\enspace x<0. \end{cases}$ Now $\arctan u=u+o(u)$, whence, if $x>0$, $$2x\arctan x=\begin{cases}\pi x-2+o\Bigl(\dfrac1x\Bigr)&\text{if}\enspace x>0,\\ -\pi x-2+o\Bigl(\dfrac1x\Bigr) &\text{if}\enspace x<0. \end{cases}$$ This proves the oblique asymptotes are \begin{align*} y&=\pi x-2\quad\text{when}\quad x\to+\infty\\ y&=-\pi x-2\quad\text{when}\quad x\to-\infty \end{align*}

Furthermore, if one uses an expansion at order $3$ for $\arctan u$, one can check the curve is above each of its asymptotes near $\infty$.