<s>[INST] Prove, that $\lim_{x \to \infty} 2^x \sin(\pi/(2^k))$ converges to $\pi$. Prove, that $$\lim_{x \to \infty} 2^x \sin\left(\frac \pi {2^x}\right)=\pi.$$
I've tried calculus to no avail.
I found out that every element of this sequence is the area of a regular rectangle with $2^{k+2}$  sides that's inside a circle whose radius is 1, therefor the function is monotonically rising and every time smaller than $\pi\cdot 1^2=\pi$.  
Is there a simpler way?
 [/INST] Using $\sin$'s MacLaurin series,
\begin{align}
\lim_{x\rightarrow\infty} \  2^x\sin\frac{\pi}{2^x} &=  \lim_{x\rightarrow\infty} \  2^x\left(\frac{\pi}{2^x} - \frac{\pi^3}{3! \cdot 2^{3x}} +  \frac{\pi^5}{5! \cdot 2^{5x} } + \ \ldots\right) \\
& = \lim_{x\rightarrow\infty}  \ \pi - \frac{\pi^3}{3! \cdot 2^{2x}} +  \frac{\pi^5}{5! \cdot 2^{4x} } + \ \ldots \\
& = \pi 
\end{align}
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