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?
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?
$1^{\rm \large st}$ 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?
$\frac{n}{\pi}\sin \frac{4\pi}{3n}=\frac{4}{3}\,\,\frac{\sin\frac{4\pi}{3n}}{\frac{4\pi}{3n}}$
\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*}