A unit circle is inscribed with a square, a circle in the square, an octagon in this circle, and again a circle in the octagon, and so on. Generally, in the $n$-th circle, a regular $2^{n+1}$-gon is inscribed, and in this, the $(n+1)$-th circle. Let $R_{n}$ be the radius of the $n$-th circle. What is $\lim _{n \rightarrow \infty} R_{n}$?
In the $n$-th circle, we inscribe a $2^{n+1}$-gon, and the central angle corresponding to one of its sides is $\frac{\pi}{2^{n}}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_cbe5deffb1252eb3b0cfg-1.jpg?height=418&width=759&top_left_y=233&top_left_x=672)

The angle bisector is perpendicular to the side, so as can be seen from the diagram:

$$
\frac{R_{n+1}}{R_{n}}=\cos \frac{\pi}{2^{n+1}}=\frac{\sin \pi / 2^{n}}{2 \sin \pi / 2^{n+1}}
$$

based on the identity $2 \cos \alpha \sin \alpha=\sin 2 \alpha$. We know that $R_{1}=1$, so

$$
R_{n}=\frac{R_{n}}{R_{n-1}} \cdot \frac{R_{n-1}}{R_{n-2}} \cdot \frac{R_{n-2}}{R_{n-3}} \cdot \cdots \cdot \frac{R_{3}}{R_{2}} \cdot \frac{R_{2}}{R_{1}}=\frac{\sin \pi / 2}{2^{n-1} \sin \pi / 2^{n}}=\frac{2}{\pi} \cdot \frac{\pi / 2^{n}}{\sin \pi / 2^{n}}
$$

Since $\lim _{n \rightarrow \infty} \frac{\pi}{2^{n}}=0$, and $\lim _{x \rightarrow 0} \frac{x}{\sin x}=1$, therefore

$$
\lim _{n \rightarrow \infty} R_{n}=\frac{2}{\pi} \cdot \lim _{n \rightarrow \infty} \frac{\pi / 2^{n}}{\sin \pi / 2^{n}}=\frac{2}{\pi}
$$

The desired limit is thus $2 / \pi$.