$ \lim_{x \to 0}x \tan (xa+ \arctan \frac{b}{x})$ I have to evaluate the following limit
$$ \lim_{x \to  0}x \tan (xa+ \arctan \frac{b}{x})$$
I tried to divide tan in $\frac{sin}{cos}$ or with Hopital but I can't understand where I'm making mistakes.
The final result is:
$\frac{b}{1-ab}$ if $ab \ne 1$
$- \infty$ if $ab=1$ and $a>0$
$+ \infty$ if $ab=1$ and $a<0$

$$\tan\left(ax+\arctan\frac bx\right)=\tan\left(ax+\frac\pi2-\arctan\frac xb\right)=\cot\left(\arctan \frac xb-ax\right)
\\\sim\dfrac1{\left(\dfrac1b-a\right)x}$$
and the limit for $ab\ne1$ is $\dfrac b{1-ab}$.
When $ab=1$, by Taylor the argument of the cotangent becomes asymptotic to $-\dfrac{a^3x^3}3$, hence the limit is $\pm\infty$, with the sign of $a$.
