Prove using integration that $polygon → circle\space \text{as}\space number\space of\space sides → infinity$ Say we have a regular polygon $s$, with number of sides $n$:
Is there a way to prove that as $n → ∞,\space $then $s → circle$ using integration?
 A: Denote length of side of a regular polygon as $s$, and the radius as $r$.  
Making a triangle for each of the $n$ sides in the polygon:
$$s / 2 = r \times \sin\left(\frac{180}{n}\right)$$
Then the perimiter of the polygon is the length of each side times the number of sides: $$s \times n = 2\times r \times \sin\left(\frac{180}{n}\right) \times n$$
Taking the limit as the number of sides goes to infinity:
$$\lim_{n \rightarrow \infty} \left[2\times r \times \sin\left(\frac{180}{n}\right) \times n\right] = 2\pi r \quad \blacksquare$$ 
For a given radius, we want to prove that as the number of sides approaches infinity, the perimeter of polygon approaches a circumference of a circle with the same radius, the two shapes are as indistinguishable as we desire to make them by drawing the polygon with n sides as close to infinity as needed.
If-then proof:
If a regular polygon has infinite sides, then it is a circle.
If the limit of the perimeter of a regular polygon approaches a circumference of 2 * Pi * r as the number of sides goes to infinity, then it is a circle.
Some would consider this "vacuous truth", because it is impossible to find a case where a polygon has infinite sides and is not a circle. The if-conditional is technically impossible, but the statement nonetheless resolves to true.
In terms of integrating, $s\times n$ is technically integrating infinitesimal sides of the same length $s$ over the perimeter of the polygon (in discrete as a summation): \Sum(x(s),{x,1,n}) = x1(s) + x2(s) + ... + xn(s) = lengthOfSide1 + lengthOfSide2 + ... = n * s. They are considered the same, akin to comparing arc length using rectification and infinitesimal calculus.
A: The pioneers of the calculus, including Leibniz, thought of a circle as an infinite-sided polygon, as the OP suggests. This turned out to be a fruitful idea, even though today we avoid this viewpoint. See this answer for aditional details.
A: Consider the equation for the measure of each an a regular polygon:
$\frac{(n-2)180^\circ}{n} $
Now, consider the function :$$f(x)= \frac{(x-2)180}{x}$$
The limit of this function as $x$ approaches infinity is $180$. 
As $x$ gets very large, the measure of the angles inside the polygon with that amount of sides gets closer and closer to $180$. 
In fact, if you had a $1000$-gon, to the naked eye it would appear a circle. However, the slight deviation would make it a polygon, not a circle, since the interior angle measures would be $179.64^\circ $ 
So this raises a question: When $x$ is infinity, do we end up with a circle? As $x$ becomes arbitrarily large, the resulting polygon would appear more and more similar to a circle, but never actually a circle. When we have an infinite number of sides, the resulting figure is not a polygon, but a circle.
When x is infinity, $f(x)$ is $180$. You might claim that a polygon with interior angle measures of 180 degrees is not possible, and you'd be right. 
In fact, this is so impossible that the resulting figure isn't a polygon; it can't be. Instead, we are left with a circle. 
So a circle is not a polygon with an infinite number of sides, because a circle isn't a polygon by any means. 
However, you should check out the apeirogon, the geometric figure with an actual infinite number of sides. 
A: The statement "as $n\to\infty,$ then $s\to\text{circle}$" is much too informal to lead at all to something that is provable.
The same goes with "$\lim_{n\to\infty}p(n)=\text{circle}$" (in the comments).
