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I am new to this forum, so I hope I am proceeding in the correct way. Please excuse any mistakes.

I am trying to prove that the limiting factor of a 2D shape's surface area is a circle, and I have managed to find an equation for this from a regular polygon. In my question, I assume that the maximum perimeter/circumference one can form is 40cm (i.e. if someone has 40cm of string). So:

For a perimeter of 40: $$ \lim_{n \rightarrow \infty}\frac{200\sin(\frac{2π}{n})}{n\sin^2(\frac{π}{n})}=\frac{400}{π} $$

I got this value using Wolfram Alpha, which confirmed that the limiting factor for surface area is a circle, since the surface area of a circle with a circumference of 40cm = 400/π.

However, I am unable to prove this formula algebraically. I tried using l'hospital's rule after realising that the limit of the original function was 0/0, but I got nowhere. In fact, the result I got was -400π/0, which was very disappointing after so much working out!

I was wondering if anyone could help me prove this algebraically or otherwise. I am happy that I found an equation that proves what I wanted to, but I am unable to prove Wolfram Alpha's result, which is frustrating.

Again, I am new to this forum, so please let me know if I have made any mistakes so that I can edit my question.

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  • $\begingroup$ l'hopital's rule will work here - are you sure you don't have a mistake in applying the chain rule to the d/dn(sin(1/n)) parts of your expression? $\endgroup$
    – jacob1729
    Oct 3 '18 at 22:13
  • $\begingroup$ Maybe - I will try again later and let you know. It's a very frustrating process though, but I'm glad it should work! $\endgroup$
    – Natumakie
    Oct 3 '18 at 22:16
  • $\begingroup$ Side note: I don't think this site is a forum. The stack exchange tour makes it clear the site is about getting answers and not irrelevant discussion $\endgroup$
    – qwr
    Oct 4 '18 at 7:05
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Let $x = \frac {\pi}{n}$

Now we have the more familiar looking:

$\lim_\limits{x\to 0} \frac {200\sin 2x}{(\frac \pi x) \sin^2 x}\\ \lim_\limits{x\to 0} \left(\frac {200}{\pi} \right)\left(\frac {\sin 2x}{\sin x}\right)\left(\frac {x}{\sin x}\right)$

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  • $\begingroup$ Imo, the accepted answer isn't clear enough, while this one is. Maybe one more line can be added: sin(2x)/sin(x) = 2sin(x)cos(x)/sin(x) = 2cos(x) $\endgroup$ Oct 4 '18 at 3:54
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It's simple if you use equivalents:

Near $0$, $\sin x\sim x$, so $$\frac{200\sin(\frac{2π}{n})}{n\sin^2(\frac{π}{n})}\sim_{n\to\infty}\frac{200\ \dfrac{2π}{n}}{n \Bigl(\dfrac{π}{n}\Bigr)^2}=\frac{\cfrac{400\not\pi}{\not n}}{\cfrac{\pi^{\not2}}{\not n}}=\frac{400}\pi.$$

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Hint: $$ \frac{200\sin(\frac{2π}{n})}{n\sin^2(\frac{π}{n})} = 200 \;\frac{\sin(\frac{2π}{n})}{\frac{2π}{n}} \left( \frac{\frac{π}{n}}{\sin(\frac{π}{n})} \right)^2 \frac{\frac{2π}{n}}{n\left(\frac{π}{n}\right)^2} $$ What can you say about $\frac{\sin(x_n)}{x_n}$ when $x_n \xrightarrow{n\to\infty} 0$?

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