Uniform convergence of $f_n(x) = \cos^n(x/ \sqrt{n})$ . I'm studying the uniform convergence of the following sequence :
\begin{equation*}
f_n(x) = 
\left\{
\begin{split}
 & \ \cos^n\left(\frac{x}{\sqrt{n}}\right) \ \ \textrm{if} \ x \in \left[0, \frac{\sqrt{n} \pi}{2} \right] \\
                           & 0 \ \ \textrm{if} \ x \geq  \frac{\sqrt{n} \pi}{2}
\end{split}
\right.
\end{equation*}
to $f : x \longmapsto e^{-x^2/2} $ .
I computed this to see that it converges uniformly but I don't get to a proof. I tried to study the difference $f - f_n$ analytically but it doesn't work.
Thanks for help. 
 A: It's generally true that for a sequence of decreasing, positive functions $(f_n)_{n\in \mathbb{N}}$ on $[0,\infty)$ pointwise convergence to a continuous function $f$ tending to $0$ as $x\to \infty$ implies uniform convergence.
To see this, note that $f$ is uniformly continuous. So let $\varepsilon>0$ be given and pick $K>0$ such that $f(x)\leq \frac{\varepsilon}{3}$ for all $x\geq K$ and pick $\delta>0$ parrying $\frac{\varepsilon}{3}$ for the uniform continuity of $f$. Let thereafter $0=t_0<t_1<t_2<...<t_J$ such that $t_J\geq K,$ and $t_{j+1}-t_j<\delta$ for every $j$.
Now, pick $N\in \mathbb{N}$ sufficiently great that $|f_n(t_j)-f(t_j)|\leq \frac{\varepsilon}{3}$ for every $j$. 
Then, for general $x\in (t_j,t_{j+1}),$ note that
$$
f_n(t_{j+1})-f(t_j) \leq f_n(x)-f(x)\leq f_n(t_j)-f(t_{j+1})
$$
implying that $|f_n(x)-f(x)|\leq \frac{2\varepsilon}{3}$. 
Similarly, for $x>K,$ we have $0\leq f_n(x)\leq f_n(t_J)\leq \frac{2\varepsilon}{3}$.
This shows uniform convergence.
A: Too long for comments.
Since composition of Taylor series is one of my hobbies, continuing your work, we have
$$\cos^n\left(\frac{x}{\sqrt{n}}\right)=e^{-\frac{x^2}{2}}\left(1-\frac{x^4}{12 n}+\frac{x^6 \left(5 x^2-32\right)}{1440 n^2}+O\left(\frac{1}{n^3}\right) \right)$$ We have also the "nice"
$$\text{sinc}^n\left(\frac{x}{\sqrt{n}}\right)=e^{-\frac{x^2}{6}}\left(1-\frac{x^4}{180 n}+\frac{x^6 \left(7 x^2-160\right)}{453600 n^2}+O\left(\frac{1}{n^3}\right) \right)$$
$$\text{tanc}^n\left(\frac{x}{\sqrt{n}}\right)=e^{+\frac{x^2}{3}}\left(1+\frac{7 x^4}{90 n}+\frac{x^6 \left(343 x^2+2480\right)}{113400 n^2}+O\left(\frac{1}{n^3}\right) \right)$$
