It is good to study a real example of approximating a circle as an infinite sided polygon.
For example, the Viete's formula:
http://en.wikipedia.org/wiki/Viète's_formula
Viete's formula represents a sequence of polygons with numbers of sides equal to [itex]2^{n}[/itex], inscribed in a circle.The Viete product is:
[itex]2/\pi = U_{1}/U_{2} \cdot U_{2}/U_{3} \cdot U_{3}/U{4} \cdot \cdot \cdot \cdot = U_{1}/U_{\infty}[/itex]
the Viete product telescopes to give the ratio of areas of a square (the initial polygon in the sequence) to a circle (the limiting case of a [itex]2^{n}[/itex]-gon).
Alternatively, the terms in the product may be instead interpreted as ratios of perimeters of the same sequence of polygons, starting with the ratio of perimeters of a digon [itex]U_{1}[/itex],(the diameter of the circle, counted twice) and a square [itex]U_{2}[/itex] , the ratio of perimeters of a square [itex]U_{2}[/itex] and an octagon [itex]U_{3}[/itex], etc etc up to the ratio of perimeters of [itex]U_{\infty-1}[/itex] and [itex]U_{\infty}[/itex] .
[itex]U_{\infty}[/itex] is the perimeter of [itex]2^{\infty}[/itex]-sided polygon. If the "radius"
of this [itex]2^{\infty}[/itex]-sided polygon is equal to 1, its diameter is equal to 2 (= [itex]U_{1}/2[/itex]), then its perimeter is equal to
[itex]2\pi[/itex], hence
[itex]U_{\infty}/(U_{1}/2) = 2U_{\infty}/U_{1}= U_{\infty}/2 = 2\pi/2 = \pi[/itex]
this is the same result as we obtained with the Viete's formula [itex]U_{1}/U_{\infty} = 2/\pi[/itex]
All the time a distinction is made between a circle and [itex]2^{\infty}[/itex]-sided polygon,
which is just the limiting case of [itex]2^{n}[/itex]-gon.
It might lead to an error to believe that a polygon transforms into a circle at an "infiniteth" step.
The error just seems to disappear if we are free to call a [itex]2^{\infty}[/itex]-sided polygon
a circle.
I used the Viete's formula from Jörg Arndt book Pi - Unleashed:
http://books.google.fi/books?id=Qww...Arndt, squaring the circle with holes&f=false