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.