Probably an old chestnut, but it was new to me

A better solution from @Brian Charlesworth , see comment below.

$\begin{aligned} L & = \lim_{n \to \infty} n\sin \frac \pi n & \small \color{#3D99F6} \text{Multiply up and down by }\frac \pi n \\ & = \lim_{n \to \infty} \frac {\pi {\color{#3D99F6}\sin \frac \pi n}}{\color{#3D99F6}\frac \pi n} & \small \color{#3D99F6} \text{Note that } \lim_{x \to 0} \frac {\sin x}x = 1 \\ & = \pi \approx \boxed{3.14} \end{aligned}$

$\begin{aligned} L & = \lim_{n \to \infty} n\sin \frac \pi n & \small \color{#3D99F6} \text{Divide up and down with }n \\ & = \lim_{n \to \infty} \frac {\sin \frac \pi n}{\frac 1n} & \small \color{#3D99F6} \text{A 0/0 cases, L'Hôpital's rule applies.} \\ & = \lim_{n \to \infty} \frac {- \frac \pi {n^2} \cos \frac \pi n}{-\frac 1{n^2}} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }n \\ & = \lim_{n \to \infty} \pi \cos \frac \pi n \\ & = \pi \cos 0 \\ & = \pi \approx \boxed{3.14} \end{aligned}$

We can also derive this formula for $π$ geometrically.

$sin(x) \approx x$ for small values of x

On the unit circle, we have $sin{\frac{\pi}{n}}$ is the height of a small right triangle with angle $\frac{\pi}{n}$ . Then, if we have $n$ of these lengths they approximate the length of the curve of the unit circle above the x axis which has length $\pi$ .