Given,

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}}$

Substituting $x = 0$ to get,

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}} = \frac{0}{0}$

As, this limit is undefined so the other approach would be,

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}}$

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}} = \frac{{\left( {\mathop {\lim }\limits_{ax \to 0} \sin \frac{{ax}}{{ax}}} \right) \times \mathop {\lim }\limits_{x \to 0} ax + \mathop {\lim }\limits_{x \to 0} bx}}{{\mathop {\lim }\limits_{x \to 0} ax + \mathop {\lim }\limits_{x \to 0} bx \times \left( {\mathop {\lim }\limits_{bx \to 0} \sin \frac{{bx}}{{bx}}} \right)}}$

Using the formula, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$ to get,

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}} = \frac{{\mathop {\lim }\limits_{x \to 0} ax + \mathop {\lim }\limits_{x \to 0} bx}}{{\mathop {\lim }\limits_{x \to 0} ax + \mathop {\lim }\limits_{x \to 0} bx}}$

\[\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}} = \frac{{\mathop {\lim }\limits_{x \to 0} \left( {ax + bx} \right)}}{{\mathop {\lim }\limits_{x \to 0} \left( {ax + bx} \right)}}\]

\[\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}} = 1\].