# $\lim_{n \to \infty}n\sin\left(\pi/n\right)$

Assume unit radius. The area of regular polygon with $n$ sides = $n\sin\left(\pi/n\right)$, how does $n\sin\left(\pi/n\right)$ approach $1$ as $n \to \infty$ ?.

$$\mbox{I have tried}\quad \pi\,{\sin\left(\pi/n\right) \over \pi/n}\quad \mbox{so it looks like}\quad \pi\,{\sin(x) \over x}$$ L'Hopital's Rule and I still can not get $\quad\lim_{x \to \infty}\sin\left(x\right)/x = 1$.

• If you're considering unit radius then the area of the regular polygon with n sides converges to the area of the circle of radius one, so it converges to $\pi$, which is what you have found. What is the problem? – Federico Oct 13 '16 at 18:31
• $\lim_{n\to \color{red}\infty}\sin(\pi/n)] / (\pi/n)=\lim_{x\to\color{red}0+}\sin x / x = 1$ – georg Oct 13 '16 at 18:33
• Of course, I don't know why i could not see that myself. Both x and sin x approaches 0 from positive. My bad...Thanks – W. Davis Oct 13 '16 at 20:37

As $\epsilon\to0$, we have $\sin\epsilon=\epsilon+o(\epsilon)$ (it's Taylor's formula around $0$, at order $1$), hence, as $n\to\infty$,

$$n\sin\frac\pi n=n\left(\frac\pi n+o(\frac1n)\right)=\pi+o(1)\to\pi$$

For small values of $x$, $\sin x\sim x$. So for large values of $x$, $\sin(1/x)\sim 1/x$. So as $n$ increases, $n\sin(\pi/n) \sim n\cdot\pi/n = \pi$.

• For $n=1000$, my calculator says that $n\sin(\pi/n)\approx3.14159$. – Mark McClure Oct 13 '16 at 18:47
• You forgot a π factor. – Bernard Oct 13 '16 at 19:01

As often, the shortest is to use equivalents:

$\sin x\sim_0 x$, hence $\sin\dfrac\pi n\sim_\infty\dfrac\pi n$ and $$n\sin\frac\pi n\sim_\infty n\dfrac\pi n=\pi.$$

• I haven't seen this notation or terminology before -- can you provide a reference? It looks like an interesting way to think about these problems which I would like to learn about – Chill2Macht Oct 13 '16 at 19:32
• It's a way to remove irrelevant details when calculating limits. $\sim_a$ reads as ‘equivalent in a neighbourhood of $a$’. You can take a look at the Wikipedia note on Asymptotic analysis. – Bernard Oct 13 '16 at 20:08

I just want to point out this can be derived from Archimedes' Method for calculating the area of a (unit) circle (here we apply it for a polygon with $2n$ sides)