Determine $\lim_{n \to \infty} (2^{n+1} \sin(\frac{\pi}{2^{n+1}}))$ I was able to calculate
$$\lim_{n \to \infty} (2^{n+1} \sin(\frac{\pi}{2^{n+1}})) = \pi$$
using L'Hopital Rule, but how to get $\pi$ without it?
 A: Make the substitution
$$\frac{\pi}{2^{n+1}} = \frac 1m\,$$
and use the limit
$$\lim_{m\to \infty}m\sin\left( \frac 1m\right) =1$$

Proof of this second notable limit. For $\alpha\in (0, \pi/2)$, we have the inequalities
$$\sin(\alpha) \le \alpha \le \tan(\alpha)\,.$$
Thus we do the following estimates
$$m\sin\left( \frac 1m\right) \le 1$$
and
$$m\sin\left( \frac 1m\right) = m\sin\left( \frac 1m\right) \frac{\cos\left( \frac 1m\right)}{\cos\left( \frac 1m\right)} = m\cos\left( \frac 1m\right)\tan\left( \frac 1m\right) \ge \cos\left( \frac 1m\right)$$
By squeeze theorem, one concludes.
A: since $\sin  \left( \frac { \pi  }{ 2^{ n+1 } }  \right) \overset { n\rightarrow \infty  }{ \longrightarrow  } 0$ we can apply famous limit  $\lim _{ n\rightarrow 0 }{ \frac { \sin { n }  }{ n }  } =1$
$$\lim _{ n\to \infty  } \left( 2^{ n+1 }\sin  \left( \frac { \pi  }{ 2^{ n+1 } }  \right)  \right) =\lim _{ n\to \infty  } \frac { \sin  \left( \frac { \pi  }{ 2^{ n+1 } }  \right)  }{ \frac { \pi  }{ 2^{ n+1 } }  } \pi =\pi $$
A: Use equivalents:
Near $0$, we have $\sin u\sim u$, so
$$2^{n+1}\sin\frac\pi{2^{n+1}}\sim_\infty2^{n+1}\frac\pi{2^{n+1}}=\pi.$$
