Limit approaching pi

Question

Consider a circle of radius 1. That circle has area pi. We can approximate that area with the area of inscribed regular polygons with n sides. By dividing the polygon in n congruent triangles we can see that the polygon has area $\frac{n*sin(\frac{2*pi}{n})}{2}$. That means that the limit of $\frac{n*sin(\frac{2*pi}{n})}{2}$ as n goes to infinity is pi. How can we show this through calculation?

Answer 1

$$\begin{align} \lim_{n\to \infty} \frac{n \sin \left(\tfrac{2\pi}{n}\right)}{2} &= \lim_{n\to \infty} \pi \cdot \frac{\sin \left(\tfrac{2\pi}{n}\right)}{\tfrac{2\pi}{n}} \\ &= \pi \cdot \lim_{\theta \to 0} \frac{\sin \theta}{\theta} \\ &= \pi \cdot 1 \\ &= \pi. \end{align}$$

Answer 2

As $x\to0$, $\sin x$ is asymptotic to $x$ (the small-angle approximation), so we have $$\lim_{n\to\infty}\frac{n\sin\frac{2\pi}n}2=\frac{n\cdot\frac{2\pi}n}2=\pi$$ (Higher-order terms in the series expansion of $\sin x$ do not affect the result here.)

Answer 3

Use the following basic limit

$$ \lim_{x\to x_0} f(x)=0\implies \lim_{x\to x_0}\frac{\sin f(x)}{f(x)}=1$$

so

$$\lim_{n\to\infty}\frac{n\sin\frac{2\pi}n}2=\lim_{n\to\infty}\,\pi\cdot\frac{\sin\frac{2\pi}n}{\frac{2\pi}n}=\pi\cdot1=\pi$$

since $\;\frac{2\pi}n\xrightarrow[n\to\infty]{}0\;$