TITLE: Having trouble showing a function is continuous: $x/(1+|x|)$
QUESTION [6 upvotes]: I am having trouble showing that a function is continuous using the $\epsilon, \delta$ definition.
My function $f:\mathbb{R}\rightarrow(-1,1)$ is defined as $f(x) = \frac{x}{1+|x|}$.
Looking at the graph, the function is clearly continuous, but when I expand I don't get anywhere:
$ |f(x) - f(x_0)| = |\frac{x}{1+|x|} - \frac{x_0}{1+|x_0|}| 
= |\frac{x(1+|x_0|) - x_0(1+|x|) }{(1+|x|)(1+|x_0|)}|$
This does not seem to simplify much at all, I cannot find a $\delta$ that satisfies:
$\forall \epsilon > 0, \exists \delta > 0, \forall x,y, \in \mathbb{R}, |x-x_0| < \delta \Rightarrow |f(x) - f(x_0)| < \epsilon $
Am I missing something here? Any advice is greatly appreciated.

REPLY [4 votes]: One way to simplify the problem is to note that $f$ is an odd function, so without loss of generality, you can assume that $x_0\in (0,\infty).$ Of course this means that $|z|=z$ for $z\in (0,\infty).$ Therefore, we have
$|f(x) - f(x_0)| = |\frac{x}{1+|x|} - \frac{x_0}{1+|x_0|}| 
= |\frac{x(1+|x_0|) - x_0(1+|x|) }{(1+|x|)(1+|x_0|)}|=|\frac{x-x_0 }{(1+|x|)(1+|x_0|)}|\le \frac{|x-x_0|}{2},$ and this implies that we may take $\delta <\min (d(x_0,0),2\epsilon)=\min (x_0,2\epsilon).$
Note: if you do not want to use the fact that $f$ is odd, you can simply repeat the above calculation for $x_0\in (-\infty,0).$ The only change you  will make is that $|z|=-z$ and this will give you the estimate $\delta<\min (|x_0|,2\epsilon).$
This leaves the proof of continuity at $x_0=0.$ In this case, you have 
$|f(x) - f(0)| = |\frac{x}{1+|x|}|\le |x|$ and you can take $\delta<\epsilon.$