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Someone showed me a derivation for the area of a circle today. They took a circle of radius $r$ and inscribed a regular polygon in the circle. If you take an $n$-sided polygon, then its area is:

$$\frac{1}{2}r^2\left(\sin{\frac{2\pi}{n}}\right)n$$

If you let $n$ go to infinity, then you get $\pi$$r^2$ as your area.

However, you are using the limit for $\sin x/x$ as $x$ goes to $0$ to derive this. In order to derive that limit, you need to show that $\sin x<x<\tan x$, which is done using the unit circle and comparing the areas of two triangles and a sector. To find the area of the sector, you need to know the area of a circle. Almost appropriately, we've reached a circular logic.

Is there any way around this?

polfosol
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Robert Fitzgerald
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  • You can use L'Hopital's rule to derive that limit, but that might be more machinery than you care for. – Kaj Hansen Sep 28 '16 at 05:45
  • How about using the Maclaurin series of the sinc function to show that its limit at zero is one? – Parcly Taxel Sep 28 '16 at 05:46
  • If I use L'Hoptial, then I am taking the derivative of sine. However, you need to know the value of the limit in order to find the derivative of sine. – Robert Fitzgerald Sep 28 '16 at 05:47
  • If you define $\sin$ in terms of the usual series, then $\pi$ is defined as twice the first strictly positive zero. It is straightforward to establish the desired result from the series. – copper.hat Sep 28 '16 at 05:51
  • To find the area of the sector, you need to know the area of a circle. That is not necessarily true. One can compare those areas just by looking at the picture and not computing the actual value of the area of sector. – polfosol Sep 28 '16 at 05:55
  • See http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 – Nick R Sep 28 '16 at 08:47

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