# Snow Flake Problem: Limit of perimeter & area at $\infty$

I am supposed to find the limits as $n\rightarrow\infty$ of the perimeter & area of a snow flake.

$$N_n = \text{Number of sides} = 3\cdot 4^n$$

$$L_n = \text{length of side} = \frac{1}{3^n}$$

$$l_n = \text{perimeter} =N_n \cdot L_n = 3(\frac{4}{3})^{n}$$

$$l_n = 4 (\frac{4}{3})^{n-1}$$

$$\lim_{n\to\infty} l_n = \lim_{n\to\infty} 4 (\frac{4}{3})^{n-1} = \infty$$

Is this correct?

For area, the link has the answer, but I don't understand why is the area given by

$$A_n = A_{n-1} + \frac{1}{4} N_n L_{n}^2 A_0$$

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The difference between $A_{n-1}$ and $A_n$ consists of a certain number of tiny equilateral triangles. What is this number, and what is the sidelength of these triangles? – Christian Blatter Mar 18 '12 at 9:56

Yes you did it right. And about area:

1. The area of an Equilateral triangle with length side A is $$A^2\frac{\sqrt{3}}{4}$$

2. For calculating length of a single side we can do this: perimeter/number of sides

so here we have $$\frac{L_n}{N_n}$$

3. The number of little triangle that add in $Nth$ step is equal to the number of (N-1)th sides: $$N_{n-1}$$

so the difference between $A_{n−1}$ and $A_n$ is: $$N_{n-1}\times(\frac{L_n}{N_n})^2\frac{\sqrt{3}}{4}$$

and we know $$N_n=4\times N_{n-1}$$

so the difference between $A_{n−1}$ and $A_n$ is: $$\frac{1}{4}\times\frac{L_n^2}{N_n}\frac{\sqrt{3}}{4}$$

Edit: But it's not as same as the link said! I think the link made a little mistake!

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