Limiting Cases Math Example 3

Follow the full solution, then compare it with the other examples linked below.

easy

For the area of a regular

n

-gon inscribed in a circle of radius

r

A_n = \frac{n}{2}r^2\sin(2\pi/n)

. What does

A_n

approach as

n \to \infty

1
As $n \to \infty$ , the $n$ -gon approaches a circle.
2
Use $\lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1$ . Let $\theta = 2\pi/n$ so $\theta \to 0$ as $n \to \infty$ .
3
$A_n = \frac{n}{2}r^2\sin(2\pi/n) = \pi r^2 \cdot \frac{\sin(2\pi/n)}{2\pi/n} \to \pi r^2 \cdot 1 = \pi r^2$ .

Answer

A_n \to \pi r^2 \text{ (area of circle)}

The limiting case

n \to \infty

recovers the circle area formula — a beautiful connection between discrete (polygon) and continuous (circle) geometry.

About Limiting Cases

Extreme values of a parameter (approaching zero, infinity, or a critical threshold) used to check formulas and reveal simplified behavior.

The formula for the sum of a geometric series [formula] ([formula]). Find the limiting case as [form

The compound interest formula is [formula]. Analyse the limiting cases [formula] (annual), [formula]

The binomial [formula] has a famous limiting case. Evaluate for [formula] with [formula] and identif