Show that $f:\mathbb{R}\to\mathbb{R}, f(x)=\frac{1}{1+|x|}$ is uniformly continuous I am supposed to use the definition of uniform continuity to show that $f:\mathbb{R}\to\mathbb{R}, f(x)=\frac{1}{1+|x|}$ is uniformly continuous.
I'd appreciate it if you could look over my proof and tell me whether it is correct and what I could do to improve it.
$$|x-x_0|\lt\delta\implies \left|\frac{1}{1+|x|}-\frac{1}{1+|x_0|}\right|\lt\epsilon$$
$$\left|\frac{1}{1+|x|}-\frac{1}{1+|x_0|}\right| = \left|\frac{1+|x|-(1+|x_0|)}{(1+|x|)(1+|x_0|)}\right| =\left|\frac{|x|-|x_0|}{1+|x|+|x_0|+|x||x_0|}\right|$$ 
$$\le \left|\frac{|x-x_0|}{1+|x|+|x_0|+|x||x_0|}\right|=
|x-x_0|\left|\frac{1}{1+|x|+|x_0|+|x||x_0|}\right| \le|x-x_0|\lt\delta $$
Let $\delta = \epsilon \implies$ f is uniformly continuous.
 A: The proof is correct (although adding slightly more prose would make it nicer).
Alternatively, if you know derivatives, you can argue in the following way:


*

*$f$ is continuous on $[-1,1]$, so it is uniformly continuous there.

*The $f'$ exists and $\lvert f'\rvert$ is bounded by $1$ in $(0,\infty)$ and $(-\infty, 0)$, so $f$ is uniformly continuous in each of these intervals.

*If a function $f$ is uniformly continuous on two intersecting intervals $I_1,I_2$, then $f$ is uniformly continuous on $I_1\cup I_2$.

*Since $f$ is uniformly continuous in each of $(-\infty,0)$, $[-1,1]$ and $(0,\infty)$, it is uniformly continuous on their union (by the preceding point).


Edit: as suggested in the comments, you can avoid considering $[-1,1]$ separately, and instead use the fact that $f'$ exists and is bounded in $(-\infty,0)$ and $(0,\infty)$ to conclude that $f$ is uniformly continuous in each of $(-\infty,0]$ and $[0,\infty)$, and then conclude as in the last point.
A: Rather than finishing the series of equalities and inequalities with $<\delta$, you should say:

given $\varepsilon>0$, we can take $\delta=\varepsilon$ and for $|x-x_0|<\delta$ we have
  $$|f(x)-f(x_0)|\le|x-x_0|<\delta=\varepsilon$$

