<s>[INST] Finding $\lim_{x \to 0} \frac {a\sin bx -b\sin ax}{x^2 \sin ax}$ 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? 
 [/INST] A good way is to use Taylor expansion. You know what is Taylor series for $sin(x)$. So, develop accordingly the different terms.  
For the numerator, you will have $\frac{1}{6} x^3 \left(a^3 b-a b^3\right)+O\left(x^4\right)$ and for the denominator $a x^3+O\left(x^4\right)$. So, the result should be $$\frac{1}{6} \left(a^2 b-b^3\right)$$ you can still simplify to $$\frac{1}{6} b (a-b) (a+b)$$
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