It's probably a known one but I've worked out this

>>7670927
I believe your limit is incorrect.

cos is between 0 and 1, so the (1- cos) factor must be either 0 or positive. If it is 0, the limit shows that pi = 0, an obvious falsehood. If (1 - cos) is positive, the limit shows that pi = infinity * (positive) = infinity, which is also obviously wrong.

>>7670932
I assume you have never taken a calculus course
lim of 0 * infinity can be finite

>>7670932
there is an n as a factor of (1-cos)

OP:
your x is wrong, you have x=2sin(alpha) or x=sqrt(2-2cos^2(alpha)).

Secondly, you need to show that the perimeter of the polygon converges towards the perimeter of a circle, which is not obvious, and in fact wrong (see fractals for example)

You can, however, use the same idea with surfaces (you can easily show convergence with the squeeze theorem).

>>7670932
cos is between -1 and 1, so 1-cos has a max at 2. Also, cos(0)=1, so the limit would be infinity*0, which is an indeterminate form and can come out to be pi.

>>7670948
also to evaluate the limit, use :
-taylor series
-l'Hôpital

>>7670948
oh you're right, i forgot the square from the cosine theorem

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>>7670952
Then again this still is a infinityx0 type limit

I guess this
> you need to show that the perimeter of the polygon converges towards the perimeter of a circle

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44KB, 720x270px

>>7670964
You never pulled out a 2 so it should be 2pi. pic related

>>7670968
Right

Does it show the work explicitly?

>>7670927
so you're trying to express [math]\pi[/math] using [math]2 \pi[/math]

>>7670981
you mean [math]\tau[/math]

>>7670981
it was more about evaluating the limit than reaching a fundamental way to express pi

>>7670997
equivalents.

sin(x) is equivalent to x as x approaches 0. so sin(2pi/n) is equivalent to 2pi/n as n goes to infinity.

And n*sin(2pi/n) is equivalent to n*2pi/n = 2pi as n approaches infinity.

>>7671006
it was cos though, not sin

>>7671029
but cos was wrong, actually was sin.

>>7670927

There's probably a better way but this works.
x = lim(n->inf, n sqrt(2-2cos(2 pi/n)))
x = lim(n->0, sqrt(2-2cos(2 pi n))/n)
x = 2 pi lim(n->0, sin(2 pi n)/sqrt(2-2cos(2 pi n))) -- l'hopitals rule
x = 4 pi^2 lim(n->0, (sin(2 pi n)/(2 pi n))/(sqrt(2-2cos(2 pi n))/n))
x = 4 pi^2 1/x
x^2 = 4 pi^2
x = 2 pi