$\lim\limits_{n \rightarrow \infty}\left[\tan \frac{\pi}{3 n}+\tan \frac{2 \pi}{3 n}+....+\tan \frac{\pi}{3}\right] \frac{1}{n}=$
Answer & explanation
Correct answer: option 1
$\lim\limits_{n \rightarrow \infty} \frac{1}{n}\left(\sum\limits_{r=1}^n \tan \left(\frac{r \pi}{3 n}\right)\right)$
$\frac{r}{n}=x, \frac{1}{n}=dx$
$\int\limits_0^{\pi / 3} \tan \left(\frac{x \pi}{3}\right) dx=\frac{3}{\pi}|\ln (\sec t)|_0^{\pi / 3}=\frac{3}{\pi} \ln 2$