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Consider $n$-sided regular polygon. Then we can calculate the area by splitting it into isosceles triangles and using the formula for triangles: $A=\frac{1}{2}ab\sin\gamma$, then we can write the area of an $n$-sided polygon as: $$A_n=n\frac{1}{2}r^2\sin{\frac{2\pi}{n}}$$ where $r$ is the radius of the excircle of the polygon. Now when we take this as $n\to\infty$ this should give us the area of the circle, which is $\pi r^2$. How does one show numerically that $$\lim_{n\to\infty}{\frac{1}{2}n\sin{\frac{2\pi}{n}}}$$ equals $\pi$. (Of course I am interested in the way without L'Hospital's rule)

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  • $\begingroup$ Do you know about little $o$ notation ? $o(t)$ ? Or $\sim$ ? In this case $\sin t \sim t $ when $t \to 0$ $\endgroup$ – stity Jan 4 '18 at 17:22
  • $\begingroup$ little $o$ notation doesn't say much to me. Your proposition that $\sin t ∼ t$ as $t\to 0$ is just as saying that $\lim_{x\to 0}{\frac{\sin{x}}{x}} = 1$ which is generally known fact. But that doesn't help me at all since I am dealing with limit of $n\to\infty$ $\endgroup$ – Michal Dvořák Jan 4 '18 at 18:01
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$$\begin{align}\lim_{n\to\infty}{\frac{1}{2}n\sin{\frac{2\pi}{n}}}&=\lim_{n\to\infty}\pi\cdot \frac{n}{2\pi} \sin\frac{2\pi}n\\ &=\lim_{t \to 0} \pi \cdot\frac{\sin t}{t}=\pi\end{align}$$ by substituting $t = \frac{2\pi}{n}$.

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