Determine if $f_n(x)= \big(\cos (\frac x {\sqrt n} )\big )^n $ converges pointwise or/and uniformly? 
Determine if $f_n(x)= \big(\cos (\frac x {\sqrt n} )\big )^n \qquad \quad$ converges pointwise or/and uniformly?

For pointwise convergence,$\forall x\in \mathbb R ,\forall \epsilon \gt0\qquad$
I should pick an $N \in \mathbb N \qquad$ such that $n\geq N \qquad$ implies $ |f_n(x)-f(x)|\lt \epsilon $
When I tried to compute $f(x)=lim_{n\to\infty}\big(\cos (\frac x {\sqrt n} )\big )^n\qquad \qquad$ ,I see(intiutively) it converges but couldn't write it properly. How should I evaluate $f(x)$ here ?
 A: For large enough $n$, 
$$
\cos(y)\sim 1-\frac{y^2}{2}+O(y^4)
$$
which furnishes us with the intuition that perhaps, in your case with $y=\frac{x}{\sqrt{n}}$ we have 
$$
\lim_{n\to \infty}f_n(x)\stackrel{?}{=}\lim_{n\to \infty}(1-\frac{x^2}{2n})^n=e^{-x^2/2}
$$
Let's prove it, 
write 
$$
\lim_{n\to \infty}\cos(x/\sqrt{n})^n=\lim_{n\to \infty}\exp(n\log\cos x/\sqrt{n})\sim \exp(\lim_{n\to \infty}n\log(1-x^2/2n+O(1/n^2))
$$
where continuity of $\exp$ was used. Next, taylor expand 
$$
\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+\cdots
$$
and note that here we have $y=O(1/n^2)-x^2/2n$ which gets small, so
$$
\log(1-x^2/2n+O(1/n^2))=O(1/n^2)-x^2/2n+\frac{(O(1/n^2)-x^2/2n)}{2}+\cdots\\
=-x^2/2n+O(1/n^2)
$$ 
so finally
$$
\lim_{n\to \infty}\cos(x/\sqrt{n})^n\sim \exp(\lim_{n\to \infty}n\log(1-x^2/2n+O(1/n^2))\\
\sim \exp(\lim_{n\to \infty}n(-\frac{x^2}{n}+O(1/n^2))\\
=\lim_{n\to \infty}\exp(-x^2/2+O(1/n))\\
=\exp(-x^2/2)
$$
Finally, to see that the convergence is not uniform, note that taking 
$$
x=2\pi\sqrt{n}
$$
we have 
$$
f_n(2\pi\sqrt{n})=1
$$
Then 
$$
\sup_{x\in \mathbb{R}}|f_n(x)-e^{-x^2/2}|\geq \sup_{x\in \mathbb{R}}|1-e^{-x^2/2}|\\
\geq |1-e^{-100}|\geq 1/2
$$
A: Because $\frac{x}{\sqrt{n}} \to 0$ there is an $N_x \in \Bbb N$ s.t. $$\cos\left(\frac{x}{\sqrt{n}}\right) > 0 \quad n \ge N_x$$
Then it holds $$\log f_n(x) = n \log\left(\cos\left(\frac{x}{\sqrt{n}}\right)\right)$$ where $\log$ denotes the natural logarithm.
By using L'Hopital's rule twice we get $$\log f_n(x) \to -\frac{x^2}{2}$$ 
And by continuity of the exponential and logarithm function it follows $$f_n(x) \to e^{-\frac{x^2}{2}}$$
pointwise.
So $f_n \to f$ pointwise with $$f(x) = e^{-\frac{x^2}{2}}$$
Because uniform convergence is equivalent by convergence w.r.t to the $\sup$ norm let's consider $$\| f_n - f \|_\infty$$ Now consider that $$\| f_n - f \|_\infty \ge |f_n\left(2\sqrt{n}\pi\right) - f\left(2\sqrt{n}\pi\right)| \ge |1 -e^{-\sqrt{n}\pi}| \ge |1 - e^{-\pi}| \ge \frac{1}{2}$$ hence we cannot have uniform convergence.
