Show that $f$($\mathbb{R}$) =($-1$,$1$). Consider the function:
$$ f: \mathbb{R}\rightarrow \mathbb{R}  $$
$$ f(x) = \frac{x}{|x|+1} $$
I already showed that this function is continuous and injective.
Maybe this could help? Maybe we can use the intermediate value theorem? The problem is that I don't have an interval [a,b] with a,b $\in$ $\mathbb{R}$. I know that I can consider the limit, but we didn't proved the fact, that I can use the intermediate value theorem for limits.
 A: Hint: First show that $|f| < 1$ (this should be pretty easy). Then take any $y \in (-1,1)$ and show that you can explicitly solve $f(x) = y$ for $x$, thereby showing that the image is $(-1,1)$ and nothing else. You might find it helpful to do $y\geq 0$ and $y<0$ separately.
A: $$\left| f(x) \right|=\left| \frac{x}{|x|+1} \right|= \frac{|x|}{|x|+1}<1$$ thus as $f(x)$ is continuos and lim as $x \rightarrow\pm\infty=\pm1$ $$-1 < f(x) < 1$$
A: Can't you just show that there exists a solution in x for any image epsilon-greater than -1, then for any image epsilon-less than 1, then show (trivially) that you can't solve for x when the image is outside of your desired interval, then use continuity and hence IVT to show that the function hits all values in $(-1+\epsilon, 1-\epsilon)$, $\forall \epsilon >0$? Seems like proving continuity was the hardest part.
A: Correct me if wrong:
1) $0\le x:$
$f(x) = \dfrac{x}{|x|+1}= 1- \dfrac{1}{x+1}.$
$x_n:= n$ , $n\in \mathbb{N}.$
$U_n:= [0,1-\dfrac{1}{n+1}],$ and 
consider 
$\bigcup_{n=1}^{\infty}U_{n}.$
$f$ is continuous, increasing, we have : $f[1,n] =U_n.$
Show that 
$\bigcup_{n=1}^{\infty}U_n = [0,1).$
Let $0 \le y \lt 1.$
There is a $n_0$ with $y \lt 1- \dfrac{1}{n_0+1} <1.$
(Details how to choose n_0 are left to do).
Then $y \in U_{n_0}$ for $0 \le y \lt 1$, I.e. 
$\bigcup_{n=1}^{\infty} U_n = [0,1).$
2) $ x \le 0$ can be treated similarly.
Comments welcome.
