limits - Finding $lim_{x to 0} frac {asin bx -bsin ax}{x^2 sin ax}$ witouth L'Hopital, what is my mistake?

I was working on this question.

$\lim_{x \to 0} \dfrac {a\sin bx -b\sin ax}{x^2 \sin ax}$

$\lim_{x \to 0} \dfrac {1}{x^2} \cdot \lim_{x \to 0} \dfrac { \frac {1}{abx}}{\frac {1}{abx}} \cdot \dfrac {a\sin bx -b\sin ax}{\sin ax}$

$\lim_{x \to 0} \dfrac {1}{x^2} \cdot \dfrac {\lim_{x \to 0} \frac {\sin bx}{bx}- \lim_{x \to 0}\frac{\sin ax}{ax}}{\frac 1b \lim_{x \to 0} \frac {\sin ax}{ax}}$

$b \lim_{x \to 0} \dfrac {1}{x^2} \cdot \dfrac {b-a}{a}$

It seems like this limit does not exist, but if you apply L'Hopital's rule you seem to get an answer. What is wrong with what I did?

Answer

One error is that
$\lim_{x \to 0} \frac{\sin cx}{cx}
= 1$, not $c$,
so your fraction is

$\frac1{x^2}\frac{1-1}{1/b}$
which is still indeterminate.