I was thinking about different ways of finding $\pi$ and stumbled upon what I'm sure is a very old method: dividing a circle of radius $r$ up into $n$ isosceles triangles each with radial side length $r$ and central angle $$\theta=\frac{360^\circ}{n}$$ Use $s$ for the side opposite to $\theta$.
Then we can approximate the circumference as $sn$. By the law of cosines: $$s=\sqrt{2r^2-2r^2 \cos{\theta}}=r\sqrt{2-2\cos{\left(\frac{360^\circ}{n}\right)}}$$
We know $\pi=\text{circumference}/\text{diameter} \approx \frac{sn}{2r}=\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$. This becomes exact at the limit:
$$\pi=\lim_{n \to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$$
Now for my question: How would you solve the opposite problem? To make my meaning more clear, above I used the definition of $\pi$ to determine a limit that gives its value. But if I had just been given the limit, what technique(s) could I have used to determine that it evaluates to $\pi$?
Best Answer
Notice, $$\lim_{n\to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{2\pi}{n}\right)}$$ $$=\lim_{n\to \infty}\frac{n}{2}\sqrt{2\left(1-\cos\left(\frac{2\pi}{n}\right)\right)}$$ $$=\lim_{n\to \infty}\frac{n}{2}\sqrt{2\left(2\sin^2\left(\frac{\pi}{n}\right)\right)}$$ $$=\lim_{n\to \infty}\frac{2n}{2}\sin\left(\frac{\pi}{n}\right)$$ $$=\lim_{n\to \infty}\frac{\sin\left(\frac{\pi}{n}\right)}{\frac{1}{n}}$$ $$=\pi\lim_{n\to \infty}\frac{\sin\left(\frac{\pi}{n}\right)}{\left(\frac{\pi}{n}\right)}$$ Let $\frac{\pi}{n}=t\implies t\to 0\ as\ n\to \infty$ $$=\pi \lim_{t\to 0}\frac{\sin t}{t}$$ $$=\pi\times 1=\pi$$