mjhjmtu

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:
$$
lim_n rightarrow inftyfrac200sin(frac2πn)nsin^2(fracπn)=frac400π
$$

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.

Let $x = frac pin$

Now we have the more familiar looking:

$lim_limitsxto 0 frac 200sin 2x(frac pi x) sin^2 x\
lim_limitsxto 0 left(frac 200pi right)left(frac sin 2xsin xright)left(frac xsin xright)$

It's simple if you use equivalents:

Near $0$, $sin xsim x$, so
$$frac200sin(frac2πn)nsin^2(fracπn)sim_ntoinftyfrac200 dfrac2πnn Bigl(dfracπnBigr)^2=fraccfrac400notpinot ncfracpi^not2not n=frac400pi.$$

Hint:
$$
frac200sin(frac2πn)nsin^2(fracπn) =
200 ;fracsin(frac2πn)frac2πn
left( fracfracπnsin(fracπn) right)^2
fracfrac2πnnleft(fracπnright)^2
$$
What can you say about $fracsin(x_n)x_n$ when $x_n xrightarrowntoinfty 0$?