0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Welcome to MSE. Please include your question in the body of the question, instead of putting it only in the title. $\endgroup$ – José Carlos Santos Sep 18 '19 at 4:30
  • $\begingroup$ Cf. this question $\endgroup$ – J. W. Tanner Sep 18 '19 at 4:33
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ (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 $\frac{\sin(x)}{x}$? $\endgroup$ – Robert Sep 18 '19 at 4:41
  • $\begingroup$ I have added more information. Is it clear now? $\endgroup$ – José Carlos Santos Sep 18 '19 at 4:43
  • $\begingroup$ Partly, but I'm still confused as to how ${\sin(x)}$ becomes $\frac{\sin(x)}{x}$ and what purpose it serves. $\endgroup$ – Robert Sep 18 '19 at 4:46
  • $\begingroup$ 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$$\lim_{x\to0}\frac{\sin(x)}x=1.$$ $\endgroup$ – José Carlos Santos Sep 18 '19 at 4:48
  • $\begingroup$ 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. $\endgroup$ – Robert Sep 18 '19 at 4:56
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.