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?

since $\sin  \left( \frac { \pi  }{ 2^{ n+1 } }  \right) \overset { n\rightarrow \infty  }{ \longrightarrow  } 0$ we can apply famous limit  $\lim _{ n\rightarrow 0 }{ \frac { \sin { n }  }{ n }  } =1$
$$\lim _{ n\to \infty  } \left( 2^{ n+1 }\sin  \left( \frac { \pi  }{ 2^{ n+1 } }  \right)  \right) =\lim _{ n\to \infty  } \frac { \sin  \left( \frac { \pi  }{ 2^{ n+1 } }  \right)  }{ \frac { \pi  }{ 2^{ n+1 } }  } \pi =\pi $$
