Firstly i confess i came across this limit trying to solve a question which has been answered in yet another manner, but I was curious and want to learn if this limit could be computed. I could have made a mistake (although i did double check before posting) or my approach may not yield due to following a red herring, but i suppose it is all part of the learning process.I was hoping some one could shed some light on this. $$\lim _{n\rightarrow \infty }\dfrac {1} {n}\left( \dfrac {\sin \dfrac {x} {2}\cos \dfrac {x} {2}\left(\dfrac {1} {n}-1\right)} {\sin \dfrac {x} {2n}}\right) $$

I suspect this should evaluate to $\dfrac{\sin x}{x}$ though i am unsure how to proceed i think the denominators are the source of my troubles. My apologies if this is trivial

  • 1
    $\begingroup$ Hint: doesn't the denominator look similar to $\sin y / y$? $\endgroup$
    – user14972
    Apr 5 '12 at 21:23

Write the denominator as

$$\frac{x\sin \frac{x}{2n}}{2\frac{x}{2n}}$$

Which goes to $\displaystyle \frac{x}{2}$ as $\displaystyle n \to \infty$.

$$\cos \left(\frac{x}{2n} - \frac{x}{2}\right) \to \cos \frac{x}{2}$$

Thus the limit as $\displaystyle n \to \infty$ is

$$ \frac{\sin x}{x}$$

  • 1
    $\begingroup$ @Hardy: You are welcome. You could have clarified this in a comment to my answer though :-) $\endgroup$
    – Aryabhata
    Apr 5 '12 at 21:47
  • $\begingroup$ Buddy i think i did understand your answer. I just came across another trignometric identity which looked slightly different than the one you quoted and for the sake of learning just tried computing with this other identity and ran into trouble and made this post. You know the rest of the story i guess. I will post my answer shortly and hopefully you could share your thoughts how your original answer and the new one i am about to post with your help too :-) are indeed very similar. I hope this would be ok with you ? $\endgroup$ Apr 5 '12 at 22:18

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.