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?
 A: $$\lim_{n\to\infty}\left(1-\frac{x^2}{2n}+o\left(\frac{1}{n}\right)\right)^n=e^{-\frac{x^2}{2}}$$
For detailed proof, see this Question about the proof of Central Limit Theorem
A: Compute the limit of the log, using Taylor's formula for the cosine at order $2$:
$$n\ln\Bigl(\cos\dfrac x{\sqrt n}\Bigr)=n\ln\biggl(1-\dfrac{x^2}{2n}+o\Bigl(\frac1n\Bigr)\biggr)=n\biggl(-\dfrac{x^2}{2n}+o\Bigl(\frac1n\Bigr)\biggr)=-\dfrac{x^2}2+o(1).$$
A: The problem boils down to proving that
$$ \lim_{n\to +\infty} n\log\cos\frac{x}{\sqrt{n}} = -\frac{x^2}{2}\tag{1} $$
and that can be achieved by squeezing. Both the functions $f(x)=\tan(x)$ and $g(x)=\frac{\tan x}{x}$ are convex on the interval $[0,1]$, hence it follows that
$$ \forall x\in[0,1],\qquad x\leq \tan(x) \leq x+x^3 \tag{2}$$
and by integrating those inequalities over the interval $[0,z]$, for some $z\in[0,1]$, we get:
$$ \forall z\in[0,1],\qquad \frac{z^2}{2}\leq -\log\cos(z) \leq \frac{z^2}{2}+\frac{z^4}{4}\tag{3} $$
so by replacing $z$ with $\frac{|x|}{\sqrt{n}}$ the claim follows by squeezing - for any $x\in\mathbb{R}$ the ratio $\frac{|x|}{\sqrt{n}}$ belongs to the interval $[0,1]$ for any $n$ large enough.
A: HINT:
$$\lim_{n \to \infty} \left( \cos\left( \frac{x}{\sqrt{n}} \right) \right)^n=\left(\lim_{n\to\infty}\left(1-\sin^2\dfrac x{\sqrt n}\right)^{-1/\sin^2\frac x{\sqrt n}}\right)^{\lim_{n\to\infty}\frac{-n\sin^2\frac x{\sqrt n}}2}$$
Now  setting $x/\sqrt n=h,$ $$\lim_{n\to\infty}\dfrac{-n\sin^2\frac x{\sqrt n}}2=-\dfrac{x^2}2\lim_{h\to0}\dfrac{\sin^2h}{h^2}=?$$
A: Alternatively $$\lim _{ n\to \infty  } \left( \cos  \left( \frac { x }{ \sqrt { n }  }  \right)  \right) ^{ n }=exp\left( \lim _{ n\to \infty  } n\ln { \left( \cos  \left( \frac { x }{ \sqrt { n }  }  \right)  \right)  }  \right) =\\ =exp\left( \lim _{ n\to \infty  } \frac { \ln { \left( \cos  \left( \frac { x }{ \sqrt { n }  }  \right)  \right)  }  }{ \frac { 1 }{ n }  }  \right) \overset { L'Hopital }{ = } exp\left( \lim _{ n\to \infty  } \frac { -\frac { \sin { \left( \frac { x }{ \sqrt { n }  }  \right)  }  }{ \cos  \left( \frac { x }{ \sqrt { n }  }  \right)  }  }{ -\frac { 1 }{ { n }^{ 2 } }  } \left( -\frac { x }{ 2n\sqrt { n }  }  \right)  \right) =\\ =exp\left( -\lim _{ n\to \infty  } \tan { \left( \frac { x }{ \sqrt { n }  }  \right)  } \frac { \sqrt { n }  }{ 2x } { x }^{ 2 } \right) ={ e }^{ -\frac{x^2}{2} }$$
A: Applying $\ln$ gives
$$\tag 1 n\cdot \frac{\ln \cos (x/\sqrt n)}{\cos (x/\sqrt n)-1}\cdot \frac{\cos (x/\sqrt n)-1}{(x/\sqrt n)^2}\cdot(x/\sqrt n)^2$$ $$= \frac{\ln \cos (x/\sqrt n)}{\cos (x/\sqrt n)-1}\cdot \frac{\cos (x/\sqrt n)-1}{(x/\sqrt n)^2}\cdot x^2.$$
Now let's use two well known limits
$$\tag 2\lim_{u\to 1} \frac{\ln u}{u-1} = 1,\,\,\lim_{u\to 0} \frac{\cos u - 1}{u^2} = -\frac{1}{2}.$$
As $n\to \infty,$ $x/\sqrt n \to 0$ and $\cos (x/\sqrt n) \to 1,$ so the limits in $(2)$ show the limit of $(1)$ is $1\cdot (-1/2)\cdot x^2 = -x^2/2.$ Exponentiating back shows the original limit is $e^{-x^2/2}.$ 
