Uniform convergence of n sin(x/n) Show that $F_n(x) = n\sin(\frac{x}{n})$ converges uniformly on $[-a,a]$ for any finite $a > 0,$ but does not converge uniformly on $\mathbb R.$
My thoughts were since $x$ is bounded by $a,$ and $n$ can be chosen large,
$\sin(\frac{x}{n})$ is about $\frac{x}{n}$ when $n$ gets large so $n\sin(\frac{x}{n})$ is about $n(\frac{x}{n}) = x.$
( limit function is $f(x) = x$)
actual proof attempt :
For all $ε>0$ & $a>0$ let $K(ε) = \frac{a}{ε}$
$\frac{a}{ε}\sin(x\frac{ε}{a})$ something and I don't know how to proceed,
(is this even the right track?)
Thanks in advance!
 A: To prove that it converges uniformly for $x \in [-a, a]$:
Let $\epsilon > 0$
$\forall x \in [-a, a], ~~ |n \sin(\frac{x}{n}) - x| \leq \max_{x \in [-a, a]} |n \sin(\frac{x}{n}) - x| \leq |n \sin(\frac{a}{n}) - a| $
As $sin(x) = x + o(x)$, $|\lim_{n \rightarrow +\infty }n \sin(\frac{a}{n}) - a| = 0 $, and so there exists $N_{\epsilon} \in \mathbb{N}$, such that:
$\forall n\geq N_{\epsilon}, ~~|n \sin(\frac{a}{n}) - a| < \epsilon $
And so, 
$\exists N_{\epsilon} \in \mathbb{N}, ~~ \forall x \in [-a, a], ~\forall n\geq N_{\epsilon}, ~~|n \sin(\frac{x}{n}) - x| < \epsilon $
Which proves that it converges uniformly in $[-a, a]$.
To prove that it is not true for $\mathbb{R}$, see that for $n$ fixed, $|n \sin(\frac{x} {n} ) | \leq n$, so $|n \sin(\frac{x} {n} ) - x|$ can be as big as you want. Therefore, it does not converge uniformly. 
A: $$\forall n >0, \forall A > 0, \exists x \in \mathbb{R}/|x-n*\sin(\frac{x}{n})|>A$$
because $$|x-n*\sin(\frac{x}{n})|\ge|\;|x|-|n*\sin(\frac{x}{n})|\;|\ge|x|-n$$
therefore
$$\implies \forall n \in \mathbb{N},  ||id-F_n||_\infty=\infty$$
