We are given,
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}},a,b \ne 0$
Substituting $x = 0$ to get,
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}} = \frac{0}{0}$
As, this limit is undefined so the other approach would be,
Now, multiplying numerator by $ax$ and multiplying denominator by $bx$ to get,
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin ax}}{{ax}} \times ax}}{{\frac{{\sin bx}}{{bx}} \times bx}}$
Using the formula, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$ to get,
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}} = \frac{a}{b}\frac{{\mathop {\lim }\limits_{ax \to 0} \frac{{\sin ax}}{{ax}}}}{{\mathop {\lim }\limits_{bx \to 0} \frac{{\sin bx}}{{bx}}}}$
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}} = \frac{a}{b} \times 1$
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}} = \frac{a}{b}$.