2022-07-08

Problem:
$\underset{n\to \mathrm{\infty }}{lim}n\cdot \mathrm{sin}\left(\frac{n}{n+1}\pi \right)=\pi$

Mekjulleymg

Expert

Note that
$n\cdot \mathrm{sin}\left(\frac{n}{n+1}\pi \right)=n\cdot \mathrm{sin}\left(\pi -\frac{1}{n+1}\pi \right)=n\cdot \mathrm{sin}\left(\frac{1}{n+1}\pi \right).$
use replacement $u=\frac{1}{n+1}$, then $u\to 0$ when $n\to \mathrm{\infty }$, also $n=\frac{1-u}{u}$. therefore
$\underset{n\to \mathrm{\infty }}{lim}n\cdot \mathrm{sin}\left(\frac{n}{n+1}\pi \right)=\underset{u\to 0}{lim}\frac{1-u}{u}\cdot \mathrm{sin}\left(\pi u\right)=\pi \underset{u\to 0}{lim}\left(1-u\right)\frac{\mathrm{sin}\left(\pi u\right)}{\pi u}=\pi \underset{u\to 0}{lim}\left(1-u\right)\cdot \underset{u\to 0}{lim}\frac{\mathrm{sin}\left(\pi u\right)}{\pi u}=\pi .$
$\underset{n\to \mathrm{\infty }}{lim}n\cdot \mathrm{sin}\left(\frac{n}{n+1}\pi \right)=\underset{u\to 0}{lim}\frac{1-u}{u}\cdot \mathrm{sin}\left(\pi u\right)=\pi \underset{u\to 0}{lim}\left(1-u\right)\frac{\mathrm{sin}\left(\pi u\right)}{\pi u}=\pi \underset{u\to 0}{lim}\left(1-u\right)\cdot \underset{u\to 0}{lim}\frac{\mathrm{sin}\left(\pi u\right)}{\pi u}=\pi .$

Do you have a similar question?