How Do You Find the Slant Asymptote of \( y = \frac{x}{2} - \tan^{-1}x \)?

[ISITIEIW]

Hey!
I know how to find slant asymptotes of regular rational functions, but what happens when the function is $y= \frac{x}{2} - \tan^{-1}x$ ?
Is there a special way to do this? I know what the $\arctan x$ function looks like and that is $y\in(-\frac{\pi}{2},\,\frac{\pi}{2})$ and it is $x\in\mathbb{R}$. The answer is (x-pi)/2

Thanks!

 

[MarkFL]

This is how I would work the problem:

First, let's analyze the derivative of the function as $x\to\pm\infty$:

\(\displaystyle \lim_{x\to\pm\infty}\frac{dy}{dx}= \lim_{x\to\pm\infty}\left(\frac{1}{2}-\frac{1}{x^2+1} \right)= \frac{1}{2}\)

Hence, the oblique asymptote will have the form:

\(\displaystyle y=\frac{1}{2}x+b\)

Now, we require the difference between this asymptote and the function to diminish to zero as $x\to\pm\infty$, so we may write:

\(\displaystyle \lim_{x\to\pm\infty}\left(\frac{1}{2}x-\tan^{-1}(x)-\frac{1}{2}x-b \right)=0\)

\(\displaystyle \lim_{x\to\pm\infty}\left(\tan^{-1}(x)+b \right)=0\)

\(\displaystyle b=\pm\frac{\pi}{2}\)

Thus, the asymptotes must be:

\(\displaystyle y=\frac{x\pm\pi}{2}\)