How can I show that a sequence of regular polygons with $n$ sides becomes more and more like a circle as $n \to \infty$? If we construct regular polygons with larger and larger numbers of sides, they will look more and more like circles.
That is intuitively true.
I hope you will help me to express and prove it mathematically.
 A: The regular polygon approaches the circle in the following sense:


*

*All vertices of the polygon are on the circle.

*The maximal distance of the polygon to the circle is given by $2R\sin^2(\frac{\pi}{2n})$, which goes to zero as $n$ goes to $\infty$.
A: It seems worth emphasizing that "look more and more like circles" admits numerous interpretations. The answers and comments currently visible say that the polygons converge to the circle in several ways: They eventually lie within arbitrarily narrow annuli just within the circle. Their areas converge to the circle's area. Their perimeters converge to the circle's circumference.  One could add more; for example, for almost all rays $R$ emanating from the origin, the direction in which the $n$-gon crosses $R$ converges to the direction in which the circle crosses $R$ (namely, perpendicular to $R$).  The "almost" here refers to the unpleasantness that a few (countably many) $R$'s pass through a vertex of one of the polygons, so the direction of crossing is undefined there, but even these $R$'s are OK if one uses the average of the directions just to the left and just to the right of $R$.  I suspect there are lots of other convergence properties that one could state and prove in this situation.  An interesting but non-mathematical question would be to determine which of the many notions of convergence cause people to say that the $n$-gons for large $n$ "look like circles".
A: Here's one serious approach: Let $f_n\colon [0,2\pi]\to \Bbb R_+$ be the function whose graph, in polar coordinates, is the regular $n$-gon centered at the origin with a vertex at $(1,0)$. Then $(f_n)$ converges uniformly to a constant function mapping any angle to $1$, whose graph is a circle.
We can also look at the limit of the area, a la Archimedes, and the perimeter.
A: There is nothing wrong with this question. All you have to do to formulate your assumption rigorously is to use the Hausdorff distance. Then you can show, that for a sequence of regular Polygons $P_n$ of the inner radius $r$ and and the circle $S^1_r$ of radius $r$ the Hausdorff distance $d_H (P_n , S^1_r ) $ tends to  $0$. More information about Regular polygons and Hausdorff distance can be found in Richard Gardner's book called "Geometric Tomography".
A: For Polygons, as number of sides increases,the length of each side is reducing
so at ∞, the length will be equal to a point.
Each Side can be assumed to be shortened to the vertex(ie a point)
Now, the polygon becomes a locus of points which are actually the vertices of the polygon.
However, each vertex is equidistant from the centre of polygon, so the resulting polygon 
becomes a locus of points, equidistant from a common centre point, which is a circle. 
After Edit: 
Now, what remains, is to prove that at n->∞, length of side tends to zero, ie a point.  
It can be proved by contradiction.
1) Suppose at n->∞, length of each side is not equal to a point, but is L.for n-> ∞, total    length = n*L = ∞.
Now, total length is infinite,however, polygon is a closed figure, so its total length is
obviously finite.
Thus this contradicts that length is infinite, which in turn contradicts our original supposition. 
Therefore, at n-> ∞, length of each side tends to 0, ie each side becomes a point. 
A: I have proven that the circle is a binary polygon and has 2^55 SIDES
The formulas are :     
      I1 is the First Increment = (2+2x) ^ (1/2).
     arc lenght =[(2-I53) ^(1/2)]*2^53=pi/2 at 90 degree.
    t= degree

   Arc length=2^54*sin(t/2^54)=pi/2 at 90 degree.

I Giuseppe Stagno the author: my email   gstagno31@gmail.com  thanks
A: The half-angle formula is sin(t/2)
S = 2*sin(t/2)       Arc length = 2^54*sin(t/2^54) = pi/2 at 90 degree
  S = 2*sin(90 degree/2) = 2^(1/2) approximate 1.4141    Arc length = pi/2 approximate 1.5707
  Sn = 2^54*sin(90 degree/2^54)   Sn = 2^53 sides
  circle total sides = 4*(2^53) = 2^55 sides
  Tanks Giuseppe Stagno
