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$

Question

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.

Answer 1

Hint:$$\lim_{x\to0}\frac{\sin(x)}{\sin(mx)}=\lim_{x\to0}\frac{\frac{\sin(x)}x}{\frac{\sin(mx)}x}=\frac1m\lim_{x\to0}\frac{\frac{\sin(x)}x}{\frac{\sin(mx)}{mx}}.$$

Answer 2

$\lim_{x\to 0} \frac{\sin(x)}{\sin(mx)} = \lim_{x\to 0} \frac{\sin(x)}{x} \frac{x}{\sin(mx)} = \frac{1}{m}\lim_{x\to 0} \frac{\sin(x)}{x} \frac{mx}{\sin(mx)} = \frac{1}{m}\lim_{x\to 0} \frac{\sin(x)}{x} \lim_{x\to 0}\frac{mx}{\sin(mx)} = \frac{1}{m}$