How to prove that this is equal to $\pi$?

Question

I was trying to prove the formula for the area of a circle (without using integrals), so I started with the $\frac{Pa}{2}$ formula and started to manipulate it until I got, for an infinite number of sides (a circle), that it's area was equal to $$r^{2}\lim_{n\to\infty}\frac{n}{tan(90°\frac{n-2}{n})}$$ (as the apothem becomes the radius, where n is the number of sides, but that is not important)

The point is, that I needed the expression after the $r^{2}$ to be equal to $\pi$ (so I got the $\pi r^{2}$ formula, which I as trying to prove). But both the numerator and denominator are infinite and I don't know how I could manipulate this to get my result. I searched in W|A but it only gave me the answer with no explanation. Could someone help?

Answer 1

Since $90^\circ=\pi/2$ and $\tan(\pi/2-a)=1/\tan(a)$, we have $$ \frac{n}{\tan(\frac\pi2\cdot \frac{n-2}{n})} = \frac{n}{\tan(\frac\pi2 - \frac\pi n)} = n\tan{\frac \pi n} $$ and for small $x$ we have $\tan x\sim x$ so the limit of this is the same as the limit of $n\frac \pi n = \pi$.

Answer 2

Hint: First you can replace $90^o$ by $\pi/2$ and you can use: $tan(\frac{\pi}{2}-x)=\frac{1}{tan(x)}$ and $\lim_{x \to 0} \frac{x}{tan(x)}=1$