# How to prove that the limit of an $n$-gon is a circle?

How does one prove that a regular $n$-gon with perimeter $1$, approaches (becomes) a circle as $n$ goes to (or if $n$ is) infinity?

It is not enough to prove that all points become equal distance to origin, since this also holds for the limiting object of the graph of largest area drawn on the square graph and enclosed by a circle as we make the squares smaller and smaller.

Is the circle the only possible object with all points distance $r$ from another, and total perimeter $2\pi r$?

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First, how do you define a limit of a sequence of polygons? – Srivatsan Dec 1 '11 at 4:25
The unique object the n-gon's approaches? – lessannoying Dec 1 '11 at 4:28
Dear qqqqqqqqqqqqqqqquqqqqqqqqqqqqq, please change your user name to something less annoying! – Mariano Suárez-Alvarez Dec 1 '11 at 7:29
It is not true that all point on the circle are at distance $r$ from another. Clearly there are points that lie very much closer together than $r$ on the circle. It is however true that all points are at a distance $r$ from a fixed point, the center. This is true only of the circle. (No need to consider the perimeter.) – Johan Jan 15 '13 at 12:47

You could define the circle as a continuous function of the interval into some space (take your pick). Then model the $n$-gon as a piecewise continuous function which agrees with the circle on $n$ points in the interval. In the limit, the set of points at which the two functions agree is dense in the interval and since they're both continuous they should agree in the target space. Though not a very geometric answer.
Here's a neat polar formula for the $n$-gon: math.stackexchange.com/questions/41940/… – dls Dec 8 '11 at 4:11
If you use it you need to show that for every $\epsilon >0$ there is a $N$ such that if $n>N$ then every point on the circle is closer than $\epsilon$ to a point on the $n$-gon and every point on the $n$-gon is closer than $\epsilon$ to a point on the circle.
This should be fairly easy. Any point on the circle is close to a vertex of the polygon if $n$ is large and any point on the polygon is also close to a vertex if $n$ is large. Since the vertices lie on both the circle and the polygon we are done.