A calculus problem by Nikola Alfredi

$\begin{aligned} L & = \lim_{n \to \infty} n \sin \left(\frac \pi n\right) & \small \blue{\text{Let }x = \frac \pi n} \\ & = \lim_{x \to 0} \pi \times \frac {\sin x}x \\ & = \boxed \pi \end{aligned}$

$Lt_{n\rightarrow \infty} (n\sin \dfrac{π}{n})=π\times Lt_{n\rightarrow \infty}\dfrac{\sin \dfrac{π}{n}}{\dfrac{π}{n}}=\boxed π$

Solution : You can think of a regular n sided polygon circumscribed by a circle of radius $\frac{1}{2}$ units , then you try to find its perimeter $P$ :

Let the side length of the polygon be $x$ units

The central angle when you join the vertices of n-gon with the centre of the circle, would be $\ \ \ \theta = \frac{2\pi}{n}$ ,

Then by using Sine-Rule :

$\frac{x}{\sin (\frac{2\pi}{n})} = \frac{\frac {1}{2}}{\sin (\frac {\pi}{2} - \frac{\pi}{n})}$

Also, $\ \ \sin (\frac{2\pi}{n}) = 2\sin (\frac{\pi}{n}) \cos (\frac{\pi}{n})\ \$ And $\ \ \sin (\frac {\pi}{2} - \frac{\pi}{n}) = \cos (\frac{\pi}{n})$

Thus $x \ = \frac{2\times \frac{1}{2} \sin (\frac{\pi}{n}) \cos (\frac{\pi}{n})}{\cos (\frac{\pi}{n})}$

Hence $x = \sin (\frac{\pi}{n}) \ \$ Now, the perimeter would be $n\times x$

So, $P = n\sin (\frac{\pi}{n})$

Thus as $\lim_{n\rightarrow\infty}\ \ \$ $P \approx 2\pi r \ \$ , here $r = \frac{1}{2}$

Thus $P \approx \pi$

So N = $lim_{n\rightarrow\infty} \ \ n\times \sin (\frac{\pi}{n}) \rightarrow \pi \ \$ Or N = $\pi$