If $\lim_{x\to 0} \frac{x^2 \sin (bx)}{ax-\sin x}=1$, then find $a,b$ $$\lim_{x\to 0} \frac{x^2 \frac{\sin bx}{bx} (bx)}{x(\frac{ax}{x}-\frac{\sin x}{x})}$$
$$=\lim_{x\to 0} \frac{bx^2 \frac{\sin bx}{bx}}{a-\frac{\sin x}{x}}$$
$$=0$$
Which is obviously wrong. I think I am suffering from lack of conceptual clarity in limits, so what exactly is going wrong in this question?

$$L=\lim_{x\to 0} \frac{x^2\sin bx}{ax-\sin x}=\lim_{x\to 0} \frac{x^2(bx-b^3x^3/6+...)}{ax-x+x^3/6}=\lim_{x\to 0} \frac{bx^3-b^5 x^5/6}{ax-x+x^3/6}$$
If $L=1$, then $a=1,b=1/6.$
