Finding $\lim_{n\to \infty} \frac {nr^2}{2}\cdot \sin \left(\frac {2\pi}{n}\right)$

Question

Consider a $n$ sided polygon inscribed in a circle of radius $r$. The area of this polygon hence can be given as $$A_n=\frac {nr^2}{2}\cdot \sin \left(\frac {2\pi}{n}\right)$$

Now consider the limit $$\lim_{n\to \infty} \frac {nr^2}{2}\cdot \sin \left(\frac {2\pi}{n}\right)$$

I know that this limit must equal $\pi r^2$ because as the number of sides tend to infinity the polygon is going to take more of the area under it and finally reaching the area of circle but I am not getting any algebraic way to prove so. But somewhere within I feel that L'Hospital theorem might be of some help.

Any hints and suggestions are welcome.

Answer 1

$A_n= ((nr^2)/2) (2π/n) \cdot \dfrac{\sin (2π/n)}{(2π/n)}.$

$\lim_{ n \rightarrow \infty } A_ n= $

$πr^2 \lim_{n \rightarrow \infty} \dfrac{\sin (2π/n)}{(2π/n)} $ where

$\lim_{n \rightarrow \infty} \dfrac{\sin(2π/n)}{(2π/n)} =1$.

Used:

$\lim_{x \rightarrow 0} \dfrac{\sin x}{x} =1.$

Answer 2

Continuing the hint by @alphacapture

As $n\to \infty$ ;$\frac {2\pi}{n}\to 0$

Let $\frac {2\pi}{n}=x$ Hence $$\lim_{n\to \infty} \frac {nr^2}{2}\cdot \sin \left(\frac {2\pi}{n}\right)=\lim_{x\to 0} \frac {\pi r^2}{x}\cdot \sin x =\pi r^2$$

Because $$\lim_{x\to 0} \frac {\sin x}{x}=1$$

Alternatively I could I have done it using L'Hospital as $$\lim_{n\to \infty} \frac {nr^2}{2}\cdot \sin \left(\frac {2\pi}{n}\right)=\lim_{n\to \infty} \frac {r^2}{\frac {2}{n}}\cdot \sin \left(\frac {2\pi}{n}\right)$$

Hence by L'Hospital theorem we have $$\lim_{n\to \infty} \frac {r^2}{\frac {2}{n}}\cdot \sin \left(\frac {2\pi}{n}\right)=\lim_{n\to \infty} \frac {-\frac {2\pi}{n^2}\cdot r^2}{\frac {-2}{n^2}}\cdot \cos \left(\frac {2\pi}{n}\right)=\lim_{n\to \infty} \pi r^2\cdot \cos \left(\frac {2\pi}{n}\right)=\pi r^2$$