Evaluate the Given limit:

$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}, a, b \neq 0$

At $x=0$, the value of the given function takes the form $\frac{0}{0}$

Now, $\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}=\lim _{x \rightarrow 0} \frac{\left(\frac{\sin a x}{a x}\right) \times a x}{\left(\frac{\sin b x}{b x}\right) \times b x}$

$=\left(\frac{a}{b}\right) \times \frac{\lim _{a x \rightarrow 0}\left(\frac{\sin a x}{a x}\right)}{\lim _{b x \rightarrow 0}\left(\frac{\sin b x}{b x}\right)} \quad\left[\begin{array}{l}x \rightarrow 0 \Rightarrow a x \rightarrow 0 \\ \text { and } x \rightarrow 0 \Rightarrow b x \rightarrow 0\end{array}\right]$

$=\left(\frac{a}{b}\right) \times \frac{1}{1}$ $\left[\lim _{y \rightarrow 0} \frac{\sin y}{y}=1\right]$

$=\frac{a}{b}$