 # Finding asymptotes of a function using limits I need to find asymptotes of a function. I didn't find any vertical asymptotes and I think there aren't any. But when trying to find a horizontal asymptote I evaluate the limit as x approaches infinity but can't successfully find it. f(x)=xarctan(x) brasocas6 2022-08-07 Answered
Finding asymptotes of a function using limits
I need to find asymptotes of a function. I didn't find any vertical asymptotes and I think there aren't any. But when trying to find a horizontal asymptote I evaluate the limit as x approaches infinity but can't successfully find it.
$f\left(x\right)=x\mathrm{arctan}\left(x\right)$
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The function f has an oblique asymptote $y=ax+b$ when $x\to \mathrm{\infty }$ iff
$\underset{x\to \mathrm{\infty }}{lim}\frac{f\left(x\right)}{x}=a$
$\underset{x\to \mathrm{\infty }}{lim}\left(f\left(x\right)-ax\right)=b$
Similar conditions hold for the case $x\to -\mathrm{\infty }$.
For $f\left(x\right)=x\mathrm{arctan}\left(x\right)$ we have when $x\to +\mathrm{\infty }$
$\frac{f\left(x\right)}{x}=\mathrm{arctan}x\to \frac{\pi }{2}$
$f\left(x\right)-\frac{\pi }{2}x=x\left(\mathrm{arctan}x-\frac{\pi }{2}\right)=-x\mathrm{arctan}\left(\frac{1}{x}\right)\to -1$
Hence there is an oblique asymptote $y=\frac{\pi }{2}x-1$ at $+\mathrm{\infty }$.
What is the asymptote when $x\to -\mathrm{\infty }$ ?