# Given $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$, Prove $\lim_{x \to 0} \frac{\sin(x)}{\sin(mx)} = \frac{1}{m}$ for $m >0$

Given $$\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1$$, prove that $$\displaystyle \lim_{x \to 0} \frac{\sin(x)}{\sin(mx)} = \frac{1}{m}$$ for $$m >0$$.

I can prove that this is true when looking at the limit from the left and right, but I have no idea how to incorporate the $$\frac{\sin x}{x}$$identity. Any assistance on how to accomplish this would be greatly appreciated.

• Welcome to MSE. Please include your question in the body of the question, instead of putting it only in the title. – José Carlos Santos Sep 18 '19 at 4:30
• – J. W. Tanner Sep 18 '19 at 4:33

Hint:$$\lim_{x\to0}\frac{\sin(x)}{\sin(mx)}=\lim_{x\to0}\frac{\frac{\sin(x)}x}{\frac{\sin(mx)}x}=\frac1m\lim_{x\to0}\frac{\frac{\sin(x)}x}{\frac{\sin(mx)}{mx}}.$$
• (new to calculus) How were you able to pull out a $\frac{1}{m}$ from the limit, and how does doing that turn ${\sin(x)}$ into $\frac{\sin(x)}{x}$? – Robert Sep 18 '19 at 4:41
• Partly, but I'm still confused as to how ${\sin(x)}$ becomes $\frac{\sin(x)}{x}$ and what purpose it serves. – Robert Sep 18 '19 at 4:46
• For any three numbers $a$, $b$, and $c$, with $b,c\neq0$, you have$$\frac ab=\frac{\frac ac}{\frac bc}.$$And the purpose is to be able to use the information that I have, which consists of$$\lim_{x\to0}\frac{\sin(x)}x=1.$$ – José Carlos Santos Sep 18 '19 at 4:48
• Awesome, I see why you did that now. I'm still a bit unsure about where the $\frac{1}{m}$ came from and what you did between the second and third step. – Robert Sep 18 '19 at 4:56
$$\lim_{x\to 0} \frac{\sin(x)}{\sin(mx)} = \lim_{x\to 0} \frac{\sin(x)}{x} \frac{x}{\sin(mx)} = \frac{1}{m}\lim_{x\to 0} \frac{\sin(x)}{x} \frac{mx}{\sin(mx)} = \frac{1}{m}\lim_{x\to 0} \frac{\sin(x)}{x} \lim_{x\to 0}\frac{mx}{\sin(mx)} = \frac{1}{m}$$