Find all values of parameters a, b such that the limit of the function is finite

From my uni's textbook:

Find all values of parameters $a$ and $b$ such that the following limit is a finite number:

$$\lim_{x \to 0}\frac{\sin(ax)-\arcsin (bx)-3x}{x^{3}}$$

The fact that $x$ is approaching $0$ is sort of pressuring me to use the fact that $\lim_{x\rightarrow 0}\frac{sin(x)}{x}= 1 $ (*)

What I've tried (and I am not sure if it's mathematically correct approach) is:

$\lim_{x\to 0}\frac{\sin (ax)-\arcsin (bx)-3x}{x^{3}}\ = \lim_{x\to 0}\frac{sin(ax).a}{a.x^{3}} - \frac{arcsin(bx).b}{b.x^{3}}-\frac{3}{x^{2}}$

Using this (*) fact, I get rid of $sin(ax)$ and $arcsin(bx)$ and :

$ \lim_{x\to 0}\frac{sin(ax).a}{a.x^{3}} - \frac{arcsin(bx).b}{b.x^{3}}-\frac{3}{x^{2}} = \lim_{x\to 0}\frac{a-b-3}{x^{2}}$

and this will converge only if $a-b-3=0$ so my answer to this problem would be that :

$a=b+3$ where both $a,b$ are real numbers.

I am not entirely sure whether I can split the fraction inside the limit (like I did) and then put it back together... I did consider using l'hospital but I got stuck once I derived the function because I think that the process of derivation led to the loss of some solutions.

Any ideas? Opinions? I would appreciate it.