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.