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I have to find its value

$$\lim \limits_{n\to\infty} \frac{n}\pi \cos \left( \frac{2\pi}{3n}\right) \sin \left( \frac{4\pi}{3n}\right)$$

Can you please give just clues for solving it?

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Do you mean $\cos\!\left(\frac{2\pi}{3n}\right)$ and $\sin\!\left(\frac{4\pi}{3n}\right)$? –  Antonio Vargas Jul 30 '12 at 1:00
    
@AntonioVargas yes –  HackToHell Jul 30 '12 at 1:00
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A hint would be the limit $\lim_{x\to 0}(\sin x)/x$. –  GEdgar Jul 30 '12 at 1:03
    
More hints: $\dfrac{\sin({1 \over n})}{{1 \over n}}.$ –  user2468 Jul 30 '12 at 1:16
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3 Answers

up vote 3 down vote accepted

$1^{\rm \large st}$ clue: What happens to the cosine term as $n\to\infty$?

$2^{\rm \large nd}$ clue: Can you relate the rest of it to $\lim\limits_{x\to0}\dfrac{\sin x}{x}$ and, if so, do you know how to do that one?

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The cos term becomes one and how do I make the limit to go to zero instead of infinity ? –  HackToHell Jul 30 '12 at 1:07
    
3rd clue: replace $n$ with $1/x$. –  Gerry Myerson Jul 30 '12 at 3:57
    
yay thanks for the help ;) –  HackToHell Jul 30 '12 at 12:31
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$$\frac{n}{\pi}\sin \frac{4\pi}{3n}=\frac{4}{3}\,\,\frac{\sin\frac{4\pi}{3n}}{\frac{4\pi}{3n}}$$

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\begin{equation*} \begin{split} \lim_{n\to\infty}\frac{n}{\pi} \cos \frac{2\pi}{3n}\sin \frac{4\pi}{3n}&= \lim_{n\to\infty} \frac{4}{3} \cos \frac{2\pi}{3n}\frac{\sin \frac{4\pi}{3n}}{\frac{4\pi}{3n}}\\ &= \frac{4}{3}.(\because \lim_{n\to\infty}\frac{\sin \frac{4\pi}{3n}}{\frac{4\pi}{3n}}=1\ \text{and} \lim_{n\to\infty}\cos \frac{2\pi}{3n}=\cos 0=1) \end{split} \end{equation*}

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Does this count as giving "just clues"? –  Gerry Myerson Jul 30 '12 at 8:50
    
You can accept this as answer! –  Kns Aug 3 '12 at 5:12
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