$ \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 $a<0$

If you perform the trigonometric expansion, you should end with
$$y=x \tan \left(a x+\tan ^{-1}\left(\frac{b}{x}\right)\right)=x\frac{b \cos (a x)+x \sin (a x)}{x \cos (a x)-b \sin (a x)}$$ and now, there is a problem because of the denominator.
Using Taylor expansions, we have
$$y=x \frac {b+ \left(a-\frac{a^2 b}{2}\right)x^2+\frac{1}{24} a^3  (a b-4)x^4+O\left(x^5\right) } { (1-a b)x+\frac{1}{6} \left(a^3 b-3 a^2\right)x^3+O\left(x^5\right) }$$
Using the long division
$$y=\frac{b}{1-a b}+\frac{a (a b (a b-3)+3)}{3 (1-a b)^2}x^2+O\left(x^4\right)$$
So, if $ab=1$ we have a poblem.
