suppose a general polygon of some side length x , as we increase the number of sides the polygon would start to look more and more like a circle , and its side length would be limit of x as x approaches 0. however is the limit was equal to 0 the there would not be any circle at all. Hence limit need to be a tiny-bit above 0

You have to define more acurately the problem. Suppose the polygon is inscribed in a circle of constant radius.

If x is the length of a side and n the number of sides, as n tends to infinity, x tends to 0. If you want to calculate the perimeter just multiplying the two limits, you get an indeterminate form "0 x infinity". The right way is to express x in term of n and the choosen radius, and you get the 2piR formula.

I'm not sure, but it looks like you may be asking how you can obtain the perimeter or area of a circle by using an infinite sequence of regular inscribed n-gons. Yes it is true that as the length of the side --->0 as n--->inf but you have to remember that 1) the distance from the vertices remains constant and will become the radius of the circle and 2) we are summing an infinite number of infinitely small sides around the edge. In the end this infinite sum converges to the perimeter of the circle.

So, if you join 2 adjacent vertices to the center you get an isosceles triangle with sides R,R and delta(s). Remember R is a constant. As n -->inf, delta(s) --> ds ( the arc of the circle) and the contained angle t --->0 since t = 2pi/n.

So ds := Rdt and P= int(ds)

P = int[rdt) from t = 0 to t = 2pi

= rt] 0 to 2pi

= 2piR

Just like 1+1/2+1/4+1/8... never actually equals 2 when we say limit we mean we can get however close we want to that value. Here too, it never actually equals zero but it can go infinitely close.

please answer if there was any mistake in this prove