Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

2 Answers

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:

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

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.