Function sequence $ \left( \cos\left( \frac{x}{\sqrt{n}} \right) \right)^n $ I'm studying about uniform convergence of function sequences. I haven't been able to prove that
$$\lim_{n \to \infty} \left( \cos\left( \frac{x}{\sqrt{n}} \right) \right)^n=e^{-\frac{x^2}{2}}.$$
Can you help me, please?

Alternatively $$\lim _{ n\to \infty  } \left( \cos  \left( \frac { x }{ \sqrt { n }  }  \right)  \right) ^{ n }=exp\left( \lim _{ n\to \infty  } n\ln { \left( \cos  \left( \frac { x }{ \sqrt { n }  }  \right)  \right)  }  \right) =\\ =exp\left( \lim _{ n\to \infty  } \frac { \ln { \left( \cos  \left( \frac { x }{ \sqrt { n }  }  \right)  \right)  }  }{ \frac { 1 }{ n }  }  \right) \overset { L'Hopital }{ = } exp\left( \lim _{ n\to \infty  } \frac { -\frac { \sin { \left( \frac { x }{ \sqrt { n }  }  \right)  }  }{ \cos  \left( \frac { x }{ \sqrt { n }  }  \right)  }  }{ -\frac { 1 }{ { n }^{ 2 } }  } \left( -\frac { x }{ 2n\sqrt { n }  }  \right)  \right) =\\ =exp\left( -\lim _{ n\to \infty  } \tan { \left( \frac { x }{ \sqrt { n }  }  \right)  } \frac { \sqrt { n }  }{ 2x } { x }^{ 2 } \right) ={ e }^{ -\frac{x^2}{2} }$$
