I was just wondering how I would find the "b" value of the slant asymptote function $y=mx+b$. I have m already which is $\frac{2}{\pi}$ for $x\to\infty$, but dont know how to find the "b" value. Any tips/hints would be appreciated.
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1See e.g. https://math.stackexchange.com/a/938325 – Minus One-Twelfth Nov 16 '23 at 21:01
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1for $x>0$, use $\tan^{-1}(x)=\pi/2-\tan^{-1}(1/x)$ and then Taylor-expand using the geometric series – Sine of the Time Nov 16 '23 at 21:45
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5Does this answer your question? Oblique asymptotes of $f(x)=\frac{x}{\arctan x}$ – Greg Martin Nov 16 '23 at 22:50
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Here's my attempt at this:
We will be letting $y=mx+b$ be the oblique asymptote as $x$ goes to $\infty$. Then we will have$$\lim_{x\to\infty}\left(\dfrac x{\arctan x}-mx-b\right)=0\implies\lim_{x\to\infty}\dfrac x{\arctan x}-mx=b$$which$$\lim_{x\to\infty}\dfrac x{\arctan x}-mx=\lim_{x\to\infty}x(1/\arctan(x)-m)=DNE\iff m\ne\dfrac 2\pi\\\because\lim_{x\to\infty}(1/\arctan x-m)=\dfrac2\pi-m\implies m=\dfrac2\pi$$We can actually now rewrite the limit as$$\lim_{x\to\infty}\dfrac{x(1-(2/\pi)\arctan x)}{\arctan x}=\lim_{x\to\infty}\dfrac2\pi\cdot x(1-(2/\pi)\arctan(x))\\=\dfrac2\pi\lim_{x\to\infty}\dfrac{1-(2/\pi)\arctan(x)}{1/x}$$And now direct substitution gives us a $\dfrac00$ indeterminate form, so we can use L'Hôptial's rule and simplify to get$$\dfrac4{\pi^2}\lim_{x\to\infty}\dfrac{x^2}{1+x^2}$$And now we can get$$\dfrac4{\pi^2}\lim_{x\to\infty}(1/x^2+1)^{-1}$$And applying direct substitution now gets us$$\dfrac4{\pi^2}(0+1)^{-1}=\dfrac4{\pi^2}$$So the $b$ you are looking for is $\frac4{\pi^2}$, we can finally write our slant asymptote function as$$\color{red}\pm\dfrac2\pi+\dfrac4{\pi^2}$$ and the reason it isn't just positive $\dfrac 2\pi$ for your $m$ is because $\dfrac x{\arctan x}$ is an even function.
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this is basically the same as this answer, linked in the comments – Sine of the Time Nov 17 '23 at 18:57
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