Asymptote of $x\arctan(x)$ In my book, the answer is:
$$y=\frac\pi2 x - 1\quad (x\to+\infty)$$
and
$$y=-\frac\pi2 x-1\quad (x\to -\infty).$$
I guess I should solve it using Taylor series, but I'm not completely sure how. 
 A: An oblique asymptote $y=mx+q$ at $\infty$ is found by computing
$$
m=\lim_{x\to\infty}\frac{f(x)}{x}
$$
which must be finite and nonzero; next you compute
$$
q=\lim_{x\to\infty}(f(x)-mx)
$$
that should be finite. Similarly at $-\infty$.
The computation for $m$ is easy:
$$
m=\lim_{x\to\infty}\frac{x\arctan x}{x}=\frac{\pi}{2}
$$
For $q$ we can do the substitution $t=\arctan x-\pi/2$, so $\arctan x=t+\pi/2$ and $x=\tan(t+\pi/2)=-\cot t$:
$$
q=\lim_{x\to\infty}x\left(\arctan x-\frac{\pi}{2}\right)=
\lim_{t\to0}(-t\cot t)=-\lim_{t\to0}\frac{t}{\sin t}\cos t=-1
$$
A: Set $x=\dfrac1u\enspace (u>0)$ first. You have:
\begin{align*}f(x)&=\frac1u\arctan \frac1u=\frac\pi{2u}-\frac1u\Bigl(u-\frac{u^3}3+o(u^3)\Bigr)=\frac\pi{2u}-1+\frac{u^2}3+o(u^2)\\
&=\frac{\pi x}2-1+\frac1{3x^2}+o\Bigl(\frac1{x^2}\Bigr)\end{align*}
Thus the asysmptote, for $x\to+\infty$, is the straight line with equation $y=\dfrac{\pi x}2-1$. Furthermore the curve is above its asymptote for $x$ big enough, due to the positive term $\;\dfrac1{3x^2}$.
For the case $x\to -\infty$, use $\;\arctan x +\arctan\dfrac1x=-\dfrac\pi2\;$ for $x<0$.
