# How to prove that the limit of this sequence is $400/\pi$

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.

• 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? Oct 3 '18 at 22:13
• Maybe - I will try again later and let you know. It's a very frustrating process though, but I'm glad it should work! Oct 3 '18 at 22:16
• 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
– qwr
Oct 4 '18 at 7:05

Let $$x = \frac {\pi}{n}$$

Now we have the more familiar looking:

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

• 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) Oct 4 '18 at 3:54

It's simple if you use equivalents:

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

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 $$x_n \xrightarrow{n\to\infty} 0$$?