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Using L'Hôpital's rule, I need to show how:

$$\lim_{n\to\infty}\frac{p^2\cot(\frac{\pi}{n})}{4n}=\pi r^2$$

Where $p$ is the perimiter of a regular polygon and $r$ is a radius. The idea of the example is to prove that an infinitely sided polygon becomes a circle.

The perimeter is fixed. The formula is actually for calculating the area of a regular polygon from its perimeter and the number of sides- therefore once $n$ reaches infinity, $p$ will become a circumference.

anon582847382
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  • Do you know the formula for the perimeter of a regular $n$-gon inscribed in a circle of radius $r$? – Daniel Fischer Jul 15 '14 at 18:27
  • I know it, but I need to be working from the perimeter here. Unless the circumradius of the polygon can be worked out. – anon582847382 Jul 15 '14 at 18:28
  • if you know L'Hopital's at 14 - you are right on track! – Alex Jul 15 '14 at 18:32
  • But the perimeter depends on $n$. Or do you keep the perimeter fixed and let the radius depend on $n$? – Daniel Fischer Jul 15 '14 at 18:33
  • One useful way of attacking this problem is expanding $\cot x$ in Taylor series around $0$, then you get the first term and the rest converge to $0$. – Alex Jul 15 '14 at 18:33
  • @DanielFischer Yes, sorry- I should have clarified. This is part of an isoperimetric problem. – anon582847382 Jul 15 '14 at 18:33
  • @Alex The thing is I don't really, I'm just getting into it. I thought this explanation might be good as part of a project I am doing. – anon582847382 Jul 15 '14 at 18:35
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    In order to use L'Hopital's Rule, you need to have a limit in the form of something that evaluates to $\frac 0 0$. Rearrange your terms to get that and try again. – Avi Jul 15 '14 at 18:40

3 Answers3

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This is an infinity/infinity form: $$ \begin{align} \lim_{n\rightarrow\infty} \frac{p^{2}\cot(\pi/n)}{4n} & = \frac{p^{2}}{4\pi}\lim_{n\rightarrow\infty}\cos(\pi/n)\frac{(\pi/n)}{\sin(\pi/n)} \\ & = \frac{p^{2}}{4\pi}\lim_{n\rightarrow\infty}\cos(\pi/n) \lim_{n\rightarrow\infty}\frac{(\pi/n)}{\sin(\pi/n)} \\ & = \frac{p^{2}}{4\pi}\lim_{l\rightarrow 0}\frac{l}{\sin(l)}=\frac{p^{2}}{4\pi}\lim_{l\rightarrow 0}\frac{1}{\cos(l)}=\frac{p^{2}}{4\pi} \end{align} $$

Disintegrating By Parts
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With $m = \pi/n$, we easily find $$\lim_{n \to \infty} \frac{\cot (\pi/n)}{n} = \lim_{m \to 0^+} \frac{m}{\pi} \cot m = \frac{1}{\pi} \lim_{m \to 0^+} \frac{m}{\tan m} = \frac{1}{\pi} \lim_{m \to 0^+} \cos m \cdot \frac{m}{\sin m}.$$ L'Hopital's rule applied to the fraction $m/\sin m$ immediately yields $1/\pi$ as the value of the limit.

heropup
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$\cot (\frac{\pi}{n}) \to_n \cot 0$, so you can expand it around $0: \cot \frac{\pi}{n} \sim \frac{n}{\pi} - \frac{\pi}{3 n} + O(n^{-3})$. Once you multiply the terms of this Maclaurin expansion by the remaining terms (I understand $p = (2 \pi r)^2$) you get the result.

This is not exactly L'Hospital's rule, but quite close.

Alex
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