Circle area and lim

Question

I was trying to show how to find $\pi$ value from formula $\pi R^2$, but I don't understand where is my mistake.

So I am calculating area using $n$ triangles 1

let $R=1$, then one triangle area is $1\cdot 1\cdot \dfrac{\sin\left(\frac{360^\circ}{n}\right)}{2}$, all plot then would be

$\displaystyle\lim_{ n\to\infty }\dfrac{n \sin\dfrac{360^\circ}{n}}{2}=\pi\cdot 1^2$

from the formula $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1$

$\displaystyle\lim_{ n\to\infty }\frac{n \sin\dfrac{360^\circ}{n}}{2}=\lim_{ n\to\infty }\dfrac{360^\circ\dfrac{n}{360^\circ} \sin \dfrac{360^\circ}{n}}{2}=180^\circ$

So I get that $180^\circ=\pi$, where is my mistake?

Answer 1

$$\lim_{n\to \infty}\frac{n\sin \frac{2\pi}{n}}{2}$$ Multiply the initial limit by $\frac{2\pi}{2\pi}$ and substitute $t=\frac{2\pi}{n}$. Note that if $n\to \infty$ then $\frac{2\pi}{n}\to 0$. $$\lim_{t\to 0}\frac{2\pi\sin t}{2t}=\pi \lim_{t\to 0}\frac{\sin t}{t}=\pi$$ This gives us our wanted answer $$\lim_{n\to \infty}\frac{n\sin \frac{2\pi}{n}}{2}=\pi$$

Answer 2

You aren't going to get the answer like this. Let's go through step by step.

Construct n equal isosceles triangles inside the circle (radius r) with vertices at the centre and touching the circumference, then the area of a triangle is $r.r sin (\theta /2) $ where $\theta$ is the angle at the centre. With n triangles $\theta = 2 \pi / n$, so that the area of a triangle is $r.r sin (\pi / n)$, and the total area of all the triangle is $n. r^2 sin (\pi / n)$.

Since we know that the area of a circle is $\pi r ^2$ we can equate this with the area of all the triangles as the number of triangles tends to infinity, i.e.

$\pi r ^2 = \lim_{n \to \infty} n. r^2 sin (\pi / n) = r^2\lim_{n \to \infty} n.sin (\pi / n) $, and so

$\pi = \lim_{n \to \infty} n.sin (\pi / n) $. This is true, but doesn't get you very far in calculating $\pi$.

Now le's look at your calculation with unit radius

(1) Your formula for the area of a single triangle $sin (360/n) /2 $ is in degrees (we'll come back to that), but is a little bit off compared to $sin (180/n) $. As n tends to infinity the two become equal so this isn't much of a problem.

(2) $\lim_{x\to0}\frac{\sin x}{x}=1$ is true, but only if x is expressed in radians (i.e. $2\pi$ radians in a circle rather than 360 degrees). If x is in degrees then the limit is $\pi / 180$. (see for example http://mathcentral.uregina.ca/QQ/database/QQ.09.08/h/jackie1.html)

(3) $\lim_{ n\to\infty }\frac{n \sin\dfrac{360}{n}}{2}=\lim_{ n\to\infty }\dfrac{360\dfrac{n}{360} \sin \dfrac{360}{n}}{2} \ne 180$ In fact, putting $360/n = \theta$ (where $\theta $ is in degrees) then $\lim_{ \theta \to 0 } 360.sin(\theta) /2 \theta = 180. \lim_{ \theta \to 0 } sin(\theta)/ \theta = 180. \pi /180 = \pi$

So, when you work through your calculation correctly in degrees, you end up with $\pi = \pi$