Evaluate $\lim_{x\rightarrow\mathrm\pi}\frac{\sin(mx)}{\sin(nx)}$ I have to evaluate $\lim_{x\rightarrow\mathrm\pi}\frac{\sin(mx)}{\sin(nx)}$ where $m,n \in\mathbb{N*}$. At first I thought I could just use the remarkable limit $\lim_{x\rightarrow0}\frac{\sin(x)}x = 1$ and the answer could just be $\frac {m}{n}$ but this is not the answer.... I mean it's a part of it but I don't understand why.
Let $y=x-\pi $. Then

$$\sin (mx)=\sin (m (y+\pi))$$
$$=(-1)^m\sin (my) \sim (-1)^mmy$$

thus, the limit is $$(-1)^{m-n} \frac {m}{n}$$
Just since $mx \to 0$ does not hold when $x\to \pi$.