\displaystyle\lim_{x\to \pi}\frac{\sin ax}{\sin bx}=\lim_{x\to \pi}\frac{(-1)^a \sin (ax-a\pi)}{(-1)^b \sin (bx-b\pi)}=\\
=(-1)^{a+b}\lim_{x\to \pi}\frac{\frac{\sin a(x-\pi)}{a(x-\pi)}}{\frac{\sin b(x-\pi)}{b(x-\pi)}}\cdot\frac{a}{b}=(-1)^{a+b}\cdot\frac{a}{b}