Geometric Approximation for Area of Circle Using Calculus

Question

A couple of years ago, I came up with this formula: $$\lim_{n\to\infty}\frac{180\left(\pi^2r^2\cot\left(\frac{180}{n}\right)\right)}{n\pi}=\pi r^2$$

I derived it from a geometric perspective and just wanted to know whether it was anything knew or if there was something like this on the web. I also wanted to know how you would derive it from a calculus perspective and whether anyone can figure how it would have been derived as I have forgotten small potions of the derivation of the formula. As an addition if you could explain, I had my original formula and had to multiply it by 180/pi or one radians, explaining the 180 on the top and the pi on the bottom. Is there a reason why I had to multiply one radian?

Thanks!

Answer 1

You can cancel $\pi r^2$ from both sides, and you're left with $$\lim_{n\to\infty}{180\over n}\cot{180\over n}=1$$ Now replace $n$ with $180/x$ and you get $$\lim_{x\to0}x\cot x=1$$ which is $$\lim_{x\to0}\cos x{x\over\sin x}=1$$ This is easy; $\cos0=1$, and it's a standard result that $\lim_{x\to0}x/\sin x=1$.

Here, $x$ is given in radians. This won't be how you found the result, but I don't see how anyone is going to be able to figure that out.

Answer 2

Assumping that $180$ means $\pi$, the formula write $$a_n=\frac{\pi ^2 r^2 \cot \left(\frac{\pi }{n}\right)}{n}$$ let $\frac \pi n=x$ to make $$a_n=\pi r^2 x \cot (x)$$ Now, using Taylor series for small values of $x$ $$x \cot (x)=x\left(\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}+O\left(x^5\right) \right)=1-\frac{x^2}{3}-\frac{x^4}{45}+O\left(x^6\right)$$ Back to $n$ $$a_n=\pi r^2 \left(1-\frac{\pi ^2}{3 n^2}-\frac{\pi ^4}{45 n^4}+O\left(\frac{1}{n^6}\right)\right)$$ which shows the limit and how it is approached. Moreover, it gives you an easy way to compute $a_n$ without the trigonometric functions.