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I came across the following problem:

Find the area of a $2^n$-gon inscribed in a circle and rigorously prove that the area tends to $\pi$ as $n\to\infty$. The area is easily shown to be $$ A=\frac{2^n}{2}\sin\left(\frac{\pi}{2^{n-1}}\right).$$ I just divided the $2^n$-gon into triangles, found the area of each triangle, and multiplied by $n$.

I've been having trouble showing that the limit tends to $\pi$, though. I tried to reduce it to using l'Hospital's rule, but I didn't get it to quite work out (I ended up with $0$, so I definitely made a mistake). If I've done something wrong anywhere along the lines, please let me know! Thanks.

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The area is not what you have written down. Maybe that's why L'H failed. – André Nicolas Dec 11 '12 at 2:37
That's a bummer. Any tips on how to improve? – Clayton Dec 11 '12 at 2:37
You just made a minor slip. I can write it out, but undoubtedly you can. – André Nicolas Dec 11 '12 at 2:38
@Clayton The area should read $$\dfrac{2^n}2 \sin \left( \dfrac{\pi}{2^{n-1}}\right)$$ since it is a $2^n$-gon and not a $n$-gon. – user17762 Dec 11 '12 at 2:39
@AndréNicolas: Thanks, you were right. I made a mental miscalculation. When I wrote it out I got it. I'll see if the limit works out now. – Clayton Dec 11 '12 at 2:43
up vote 1 down vote accepted

Hint: using the $A$ given in the comments, we can avoid using L'Hopital using a well known limit, think about this:

$$A = \frac{\sin(\frac{\pi}{2^{n-1}})}{\frac{\pi}{2^{n-1}}} \times \frac{2^n\pi}{2^n}$$

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As stated above in the comments, the area can be calculated to be $$ A=2^{n-1}\sin\left(\frac{\pi}{2^{n-1}}\right).$$ With that in mind, we can replace $n$ by $x$ and consider the limit as $x\to\infty$. By l'Hospital's, this gives $$\begin{align}\lim_{x\to\infty}\frac{\sin\left(\frac{\pi}{2^{x-1}}\right)}{2^{1-x}}&=\lim_{x\to\infty}\frac{\cos\left(\frac{\pi}{2^{x-1}}\right)\cdot\pi\cdot\left(\frac{1}{2}\right)^{x-1}\log(\frac{1}{2})}{2^{1-x}\cdot\log(2)\cdot(-1)}\\ \\ &=\pi\end{align}$$ after all of the cancellations.

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