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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.

Blue
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Andrei Stan
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    The sine function is zero at integer multiples of $\pi$ so you get $\frac{0}{0}$ which is indeterminate. Using L'Hospital's rule gives the result. – John Douma Nov 03 '21 at 23:10
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    Can you use Taylor expansions to solve this? – David Nov 03 '21 at 23:15
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    It will be clear if you graph both $\sin(mx)$ and $\sin(nx)$ near $x = \pi$. You will notice that $\sin(mx)$ looks almost like $(-1)^mm(x - \pi)$ near $x = \pi$, while $\cos(mx)$ looks almost like $(-1)^nn(x - \pi)$. Try to explain these by taking derivatives. – WhatsUp Nov 03 '21 at 23:20

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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 $$ (-1)^m(-1)^{-n}\frac{m}{n}=(-1)^{m-n}\frac{m}{n} $$

egreg
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