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What is the limit of the area of an n-sided polygon, as n approaches infinity?

Is it essentially the same as the area of a circle with radius $r$, i.e $\pi r^2$?

Or am I mistaken?

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    What is $r$?$\phantom{}$ – 2017-01-04
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    You mean, regular polygon? Inscribed in a circle of radius $r$? – 2017-01-04
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    If r is the distance from center to vertex, then yes you are right. – 2017-01-04
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    If regular, it will resemble a circle. But as you did not specify, if it is irregular, the circle is the upper bound for the area, and it could be anything less than that. – 2017-01-04
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    @DougM You are right, if the polygon is inscribed in a circle of radius $r$, but this does not hold if the polygon is circumscribed around the circle. – 2017-01-04
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    @pseudoeuclidean if you are going to do it right, you do it the way that Archimedes did it. There is a circle inscribed and a circle that circumscribes. As $n$ gets large, the radii of the two circle converges. – 2017-01-04
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    @close voters No comment or suggestion before you voted?? – 2017-01-06

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For a $n$-sided polygon, you can divide the polygon into $n$ triangles by joining center to vertices. You can give its area as:

$$A_n=\frac{n}{2}r^2\sin(\frac{2\pi}{n})$$

where $r$ is distance from center to a vertex.

As $n\to\infty$;

$$\lim_{n\to\infty}A_n=\pi \cdot r^2\cdot\frac{\sin(\frac{2\pi}{n})}{\frac{2\pi}{n}}= \pi r^2$$

Since $\lim_{x\to\infty}\frac{\sin x}{x}=1$.

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    $r$ may also be the distance from the center to the midpoint of each side (or any point on each side, since those points converge). – 2017-01-04
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    @pseudoeuclidean That is also true. Then the area in that case area would be $A_n=n\cdot r^2 tan(\frac{\pi}{n})$. – 2017-01-04
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    Does this formula fit for irregular polygons? – 2017-01-04
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    @TheBitByte No. I assumed it to be regular polygon. – 2017-01-04
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    What would it converge to in that case? – 2017-01-04