# 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.

• @Tolaso No, you can't... I mean I don't think, you'll get something with -1 and I don't know if it's really okay. Commented Apr 12, 2018 at 11:17
• Yes, you can! See my answer. Commented Apr 12, 2018 at 11:24
• Did you tag the question with [limits-without-lhopital] because that method must not be used, or because you think that it cannot be used here? Commented Apr 12, 2018 at 11:29
• @MartinR because I shouldn't use it to solve this Commented Apr 12, 2018 at 11:34

Let $y=x-\pi$. Then

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

thus, the limit is $$(-1)^{m-n} \frac {m}{n}$$

HINT

Let $x=y+\pi$ then

$$\lim_{x\rightarrow\mathrm\pi}\frac{\sin(mx)}{\sin(nx)}=\lim_{y\to0}\frac{\sin(m\pi+my)}{\sin(n\pi+ny)}$$

then consider the cases with $m,n$ odd or even.

Following my suggestion we can invoke DeL' Hospital's rule. The limit is of the form $\frac{0}{0}$ , hence

$$\lim_{x\rightarrow \pi} \frac{\sin mx}{\sin nx} = \lim_{x \rightarrow \pi} \frac{m\cos mx}{n\cos nx} = \frac{m}{n}\lim_{x \rightarrow \pi} \frac{\cos mx}{\cos nx} = (-1)^{m-n} \frac{m}{n}$$

because $\cos n \pi = (-1)^n$. You can prove that inductively.

Update: I have not seen that the question is tagged as "limit without DeL' Hospital". If moderators, judge , necessary please remove this answer.

• Only that the question is tagged [limits-without-lhopital] ... Commented Apr 12, 2018 at 11:25
• @MartinR Whoops, I did not see that in the first place! Should I delete the answer? Commented Apr 12, 2018 at 11:25
• It might be that the question was tagged with [limits-without-lhopital] only because OP thinks that that method does not work here (i.e. an XY problem). Perhaps OP should clarify. Commented Apr 12, 2018 at 11:28
• @Tolaso It's okay for me. I like to see more answer and I probably shouldn't tag it as [limits-without-lhospital] but I wanted a different way rather than L'Hospital but it's okay. Commented Apr 12, 2018 at 11:28

Let $$x=\pi -y$$.

$$\lim_{x\rightarrow \pi }\frac{\sin\left ( nx \right )}{\sin\left ( mx \right )}\\ \\=\lim_{y\rightarrow 0}\frac{\sin\left [ n\left ( \pi -y \right ) \right ]}{\sin\left [ m\left ( \pi -y \right ) \right ]}\\ \\=\lim_{y\rightarrow 0}\frac{\sin\left ( n\pi -ny \right )}{\sin\left ( m\pi -my \right )}\\ \\ =\frac{\cos \left ( n\pi \right )}{\cos \left ( m\pi \right )}\lim_{y\rightarrow 0}=\frac{\sin\left ( ny \right )}{\sin\left ( my \right )}\\ \\=\lim_{y\rightarrow 0}\frac{\sin\left ( n\pi \right )\cos\left ( ny \right )-\cos\left ( n\pi \right )\sin\left ( ny \right )}{\sin\left ( m\pi \right ) \cos\left ( my \right )-\cos\left ( m\pi \right )\sin\left ( my \right )}\\\\=\lim_{y\rightarrow 0}\frac{-\cos\left ( n\pi \right )\sin\left ( ny \right )}{-\cos\left ( m\pi \right )\sin\left ( my \right )} \\\\=\frac{\cos \left ( n\pi \right )}{\cos \left ( m\pi \right )}\lim_{y\rightarrow 0}\frac{n\cos\left ( ny \right )}{m\cos\left ( my \right )}\\ \\=\frac{\cos \left ( n\pi \right )}{\cos \left ( m\pi \right )}\frac{n}{m}\lim_{y\rightarrow 0}\frac{\cos\left ( ny \right )}{\cos\left ( my \right )}\\ \\=\frac{\cos \left ( n\pi \right )}{\cos \left ( m\pi \right )}\frac{n}{m}$$

If $$m$$ and $$n$$ are both even or both odd, $$\frac{\cos \left ( n\pi \right )}{\cos \left ( m\pi \right )}\frac{n}{m}=1\cdot \frac{n}{m}=\frac{n}{m}$$

If one is even and the other is odd, $$\frac{\cos \left ( n\pi \right )}{\cos \left ( m\pi \right )}\frac{n}{m}=\left ( -1 \right )\cdot \frac{n}{m}=-\frac{n}{m}$$

Just since $mx \to 0$ does not hold when $x\to \pi$.