A geometric determination of $\displaystyle \lim_{x\to 0}\frac{\sin x}{x}$ using inscribed polygons

We present a short geometric argument to determine the value of \( \lim_{x \to 0} \frac{\sin x}{x} \), based on the convergence of areas of regular polygons inscribed in the unit circle.

---

Proof.

Assume that the limit exists and denote \[ \lim_{x \to 0} \frac{\sin x}{x} = a, \] where \(a\) is a real constant to be determined.

Set \[ x = \frac{2\pi}{n}, \] so that \(x \to 0\) as \(n \to \infty\). Then \[ \lim_{n \to \infty} \frac{\sin\left(\frac{2\pi}{n}\right)}{\frac{2\pi}{n}} = a. \]

Multiplying both sides by \(\pi\), we obtain \[ \lim_{n \to \infty} \frac{n}{2} \sin\left(\frac{2\pi}{n}\right) = a\pi. \]

Observe that \[ A_n = \frac{n}{2} \sin\left(\frac{2\pi}{n}\right) \] is exactly the area of a regular \(n\)-gon inscribed in the unit circle of radius 1. Indeed, the polygon consists of \(n\) congruent isosceles triangles, each with central angle \(2\pi/n\) and area \(\tfrac{1}{2}\sin(2\pi/n)\).

As \(n \to \infty\), the inscribed polygon converges to the unit circle. Therefore, \[ \lim_{n \to \infty} A_n = \pi, \] the area of the unit circle.

Hence, \[ a\pi = \pi, \] which implies \[ a = 1. \]

Thus, \[ \lim_{x \to 0} \frac{\sin x}{x} = 1. \]