TITLE: Why does $ \lim_{x \rightarrow \pi }\frac{\sin(mx)}{\sin(nx)}=\left ( -1 \right )^{m-n}\;\frac{m}{n}$, for positive naturals $m$ and $n$?
QUESTION [5 upvotes]: Encountered this in a sample university admission exam.
$$
\lim_{x \rightarrow \pi } \frac{\sin(mx)}{\sin(nx)} \quad n,m\in \mathbb N_{> 0}
$$
What surprised me was that the answare sheet suggested that the limit was equal to:
$$
\left ( -1 \right )^{m-n}\;\frac{m}{n}
$$
Graphing the function made it clear for me that this is the correct answer, but i cannot understand why.

REPLY [5 votes]: You can substitute $x=y+\pi$, so you get
$$
\lim_{y\to0}\frac{\sin(my+m\pi)}{\sin(ny+n\pi)}
$$
Since $\sin(\alpha+k\pi)=(-1)^k\sin\alpha$, you get
$$
\lim_{y\to0}\frac{(-1)^m\sin(my)}{my}\frac{ny}{(-1)^n\sin(ny)}\frac{m}{n}
$$
and you can apply the basic limit
$$
\lim_{x\to0}\frac{\sin x}{x}=1
$$
so you get
$$
(-1)^m(-1)^{-n}\frac{m}{n}=(-1)^{m-n}\frac{m}{n}
$$