Uniform convergence of $f_n(x) = n \sin(\frac{x}{n}) , x \in [-r,r]$

Question

It is asked to prove that $$f_n(x) = n \sin(\frac{x}{n}) , x \in [-r,r]$$

Converges uniformly on the given interval for $r>0.$

The resolution of this suggested considered the fact that the function

$$x-n\sin(\frac{x}{n}) $$

is increasing on $[0, \infty)$. In particular, it is on $[0, r]$ and, $\forall x \in [0,r] $

$$0 \leq x - n\sin(\frac{x}{n}) \leq r - n\sin(\frac{r}{n}) $$

Since

$$\lim_{n \to \infty} r -n\sin(\frac{r}{n}) =0$$

$\forall \varepsilon >0, \exists n_0$ such that

$$n \geq n_0 \Rightarrow \left|r -n\sin(\frac{r}{n}) \right| < \varepsilon$$

But with this step, the author says that $\forall x \in [-r,r]$

$$n \geq n_0 \Rightarrow \left|x-n\sin(\frac{x}{n}) \right| < \varepsilon$$

I dont understand how he extended for $[-r, 0]$

Thanks in advance!

@Edit:

If someone want to give a better proof of this, be welcome, but I would like to understand this one too.

Answer 1

Note that if $x\in[0,r]$, then $$\left|x-\sin\left(\frac xn\right)\right|=\left|-x-\sin\left(\frac {-x}n\right)\right|$$

That is, the values in $[-r,0]$ are the same as in $[0,r]$.

Answer 2

Note that $0\leq x-\sin x\leq\frac{1}{6}x^3$ for $x\geq0$, so $$|f_n(x)-x|=n|\frac{x}{n}-\sin\frac{x}{n}|\leq n\cdot\frac{1}{6}|\frac{x}{n}|^3\leq\frac{r^3}{6n^2}$$

Answer 3

For every $a>0$: $$ a-\sin a=\int_0^a(1-\cos t)\,dt=2\int_0^a\sin^2 \Big(\frac{t}{2}\Big)\,dt\le 2\int_0^a\Big(\frac{t}{2}\Big)^2\,dt=\frac{a^3}{6} $$ and hence, for every $x\in \mathbb R$ $$ \lvert x-\sin x\rvert\le\frac{\lvert x\rvert^3}{6}. $$ thus for $x\in [-r,r]$: $$ \left|\,x-n\sin\left(\frac{x}{n}\right)\right|=n\left|\frac{x}{n}-\sin\left(\frac{x}{n}\right)\right|\le n\cdot \frac{x^3}{6n^3}=\frac{r^3}{6n^2}\to 0, $$ uniformly in $x$.