limits - Function sequence $ left( cosleft( 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?

Answer

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