Does a regular polygon's area approach circle area as n approaches infinity?

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John Clement Husain
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Does the limit as n approaches infinity of the area of an n-sided polygon equal to the area of a circle?
Hello everyone!
I have been looking for a general equation for any regular polygon and I have arrived at this equation:

$$\frac{nx^{2}}{4}tan(90-\frac{180}{n})$$

Where x is the side length and n the number of sides.

So I thought to myself "if the number of sides is increased as to almost look like a circle, does it result in the area of a circle?"

Is this:

$$\lim_{n\to\infty} \frac{nx^{2}}{4}tan(90-\frac{180}{n}) = \frac{C^{2}}{4\pi}$$

true?
 
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$$tan(90-\frac{180}{n}) $$
should become smaller with increasing n, to get a finite limit.
 
John Clement Husain said:
Summary: Does the limit as n approaches infinity of the area of an n-sided polygon equal to the area of a circle?

Hello everyone!
I have been looking for a general equation for any regular polygon and I have arrived at this equation:

$$\frac{nx^{2}}{4}tan(90-\frac{180}{n})$$

Where x is the side length and n the number of sides.

So I thought to myself "if the number of sides is increased as to almost look like a circle, does it result in the area of a circle?"

Is this:

$$\lim_{x\to\infty} \frac{nx^{2}}{4}tan(90-\frac{180}{n}) = \frac{C^{2}}{4\pi}$$

true?

Why are you taking the limit as ##x \rightarrow \infty##?

If you keep ##x## fixed, then the area is infinite as ##n \rightarrow \infty##. If you want your polygon to tend to a finite shape, then you need ##x## and ##n## to be related.

Note that using ##l## for the length of a side might have been more conventional.

It might be simpler to look at the angle, ##\theta = 2\pi / n## at the centre of the polygon and have a fixed distance to the vertices, ##r##, say.

Then you let ##n \rightarrow \infty## and see what happens to the limit of the area of the polygons. Note that the length of the sides of the polygon will tend to ##0## in this case.
 
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I extended the title to more accurately describe the topic.
 
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PS what's perhaps more interesting is to show that the length of the perimeter of the polygons, ##nl##, tends to ##2\pi r##.
 
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PeroK said:
Why are you taking the limit as ##x \rightarrow \infty##?

Oh! right, that was a typo.[/QUOTE]
 
John Clement Husain said:
Oh! right, that was a typo.

You still need to calculate ##x## in terms of ##n##.
 
It's easier to see if you find the area in terms of one variable relating to a circle. Take for instance the area of a regular polygon in terms of the circumradius r and number of sides n.

##A = \frac{r^2 n sin(\frac{2\pi}{n})}{2}##

The limit of this formula as the number of sides ##n\rightarrow\infty## is the familiar formula for the area of a circle.