The Ratio of the Areas

Consider an isosceles triangle formed by one side of the unit regular $n$ -gon with the center of the $n$ -gon. Let the circumradius and inradius of the $n$ -gon be $R$ and $r$ respectively. By Pythagorean theorem, we have $R^2 = r^2 + \left(\frac 12\right)^2 = r^2 + \frac 14$ . We note that the angle of of the isosceles triangle at the circle of the $n$ -gon is $\frac {2\pi}n$ , then $r = \frac 1{2\tan \frac \pi n}$ . Then we have:

$\begin{aligned} L & = \lim_{n \to \infty} \frac {A(n)-I(n)}{C(n) - A(n)} \\ & = \lim_{n \to \infty} \frac {\frac {nr}2 - \pi r^2}{\pi \left(r^2 + \frac 14\right) - \frac {nr}2} \\ & = \lim_{n \to \infty} \frac {\frac n{4\tan \frac \pi n} - \frac \pi{4\tan^2 \frac \pi n}}{\frac \pi{4\tan^2 \frac \pi n} + \frac \pi 4 - \frac n{4\tan \frac \pi n}} & \small \color{#3D99F6} \text{Multiply up and down by }4\tan^2 \frac \pi n \\ & = \lim_{n \to \infty} \frac {n \tan \frac \pi n - \pi}{\pi + \pi \tan^2 \frac \pi n - n \tan \frac \pi n} & \small \color{#3D99F6} \text{Divide up and down by }n \\ & = \lim_{n \to \infty} \frac {\tan \frac \pi n - \frac \pi n}{\frac \pi n + \frac \pi n\tan^2 \frac \pi n - \tan \frac \pi n} & \small \color{#3D99F6} \text{Let }x = \frac \pi n \\ & = \lim_{x \to 0} \frac {\tan x - x}{x + x \tan^2 x - \tan x} \\ & = \lim_{x \to 0} \left(\frac {x + x \tan^2 x - \tan x}{\tan x - x}\right)^{-1} \\ & = \lim_{x \to 0} \left({\color{#3D99F6}\frac {x \tan^2 x}{\tan x - x}} -1 \right)^{-1} & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \left({\color{#3D99F6}\frac {\tan^2 x + 2x\tan x \sec^2 x}{\sec^2 x - 1}} -1 \right)^{-1} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \lim_{x \to 0} \left(\frac {\tan^2 x + 2x\tan x \sec^2 x}{\tan^2 x} -1 \right)^{-1} \\ & = \lim_{x \to 0} \left(1 + {\color{#3D99F6}\frac {2x}{\tan x}} + \tan x -1 \right)^{-1} & \small \color{#3D99F6} \text{A 0/0 case again} \\ & = \lim_{x \to 0} \left({\color{#3D99F6} \frac 2{\sec^2 x}}\right)^{-1} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \frac 12 = \boxed{0.5} \end{aligned}$