Comparing between regular polygons

Consider that the apothem $a$, circumcircle radius $c$, and edge of the polygon form a right triangle.

That is, we have two legs, one of length $1/2$, the other of length $a$, and a hypotenuse of length $c$. Then we have:

$$a = \frac{\tan\left(\frac{(n-2)\pi}{2n}\right)}{2}$$

$$c = \frac{1}{2}\csc\left(\frac{\pi}{n}\right)$$

Since the angle $\angle ac$ is always $\pi/n$.

We may simplify $a + c = \frac{1}{2}\cot\left(\frac{\pi}{2n}\right)$. Then what you seek to compute is:

$$\lim_{n \rightarrow \infty} \frac{1}{2}\left(\cot\left(\frac{\pi}{2n}\right) - \cot\left(\frac{\pi}{2n-2}\right)\right)$$

This in fact converges to $1/\pi$, which is $\approx 0.318$.