$\begin{align}
\lim_{x \to \pi} \frac{\sin a x}{\sin b x}
& = \left[
\begin{array}{ccc}
t = x - \pi & \Rightarrow & x = t + \pi \\
x \to \pi & \Rightarrow & t \to 0
\end{array}\right]
= \lim_{t \to 0} \frac{\sin(a t + a \pi)}{\sin(b t + b \pi)}
= \lim_{t \to 0} \frac{\sin a t \cos a \pi + \cos a t \sin a \pi}{\sin b t \cos b \pi + \cos b t \sin b \pi} \\
& = \lim_{t \to 0} \frac{\sin a t \, (-1)^a}{\sin b t \, (-1)^b}
= (-1)^{a - b} \lim_{t \to 0} \frac{\sin a t}{a t} \cdot \frac{b t}{\sin b t} \cdot \frac{a}{b}
= (-1)^{a - b} \cdot \frac{a}{b}
\end{align}$