# Thread: Limit and area of a circle

1. ## Limit and area of a circle

I have the following expression:

How do you show that:

Thanks!

2. If you rewrite it a little, it becomes familiar.

$\frac{1}{2} \cdot n \cdot r^{2} \cdot \sin\left(\frac{2 \cdot \pi}{n}\right)$

$\frac{1}{2} \cdot r^{2} \cdot \frac{\sin\left(\frac{2 \cdot \pi}{n}\right)}{\frac{1}{n}}$

$r^{2} \cdot \frac{\sin\left(\frac{2 \cdot \pi}{n}\right)}{\frac{2}{n}}$

$\pi \cdot r^{2} \cdot \frac{\sin\left(\frac{2 \cdot \pi}{n}\right)}{\frac{2 \cdot \pi}{n}}$

3. Originally Posted by ty2391
I have the following expression:

How do you show that:

Thanks!

For the following proof you need to know that $\lim_{x\to 0}\frac{sin x}{x}=1$, and using continuity or L'Hospital's rule, we can prove then that in fact $\lim_{h(n)\to 0}\frac{sinh(n)}{h(n)}$ if $h(n)$ is a sequence converging to zero, so:

$\lim_{n\to 0}\frac{1}{2}nr^2sin\frac{2\pi}{n}=\lim_{n\to 0}\frac{2\pi r^2}{2}\,\frac{sin\frac{2\pi}{n}}{\frac{2\pi}{n}}= \pi r^2\cdot 1=\pi r^2$

Tonio