How to determine the range of the following function $\frac{x}{1+ |x|}$? How to determine the range of the following function $\frac{x}{1+ |x|}$?
when I calculated it, it was $\mathbb{R}$, but my professor said that the range is ]-1,1[, could anyone explain for me why?
thanks!  
 A: Let $f(x)=\frac{x}{1+ |x|}$. Then:  $|f(x)|=\frac{|x|}{1+ |x|} \le 1$, hence
$f( \mathbb R) \subseteq [-1,1]$.
Furthermore: $\lim_{x \to \infty}f(x)=1$  and $\lim_{x \to -\infty}f(x)=-1$.
Show that $f(x) \ne 1$ and $f(x) \ne -1$ for all $x$.
Are you now in a position to derive $f( \mathbb R) =]-1,1[$ ?
A: For $x\geq 0$, $\dfrac{x}{1+|x|}=\dfrac{x}{1+x}=1-\dfrac{1}{1+x}$, it is increasing on $[0,\infty)$, so it maps onto $[0,1)$. Similarly you can deal with $(-\infty,0]$.
A: check : $1+|x|\neq 0\Rightarrow |x|\neq-1$, which is true for all $x$.
Now,
when $0\leq x<\infty , y=\frac{x}{1+x}, x<x+1\Rightarrow 0\leq y<1...(I)$
when $-\infty <x<0 , y=\frac{x}{1-x}$ , 
|numerator|<|denominator|, so $|y|$ lies between $0$ and $1$, but since numerator is negative and denominator is positive,$-1<y<0...(II)$
combining $(I),(II)$ we get $-1<y<1$
A: Observe that $\large|\frac {x}{1+|x|}|=\frac {|x|}{1+|x|}=1-\frac {1}{1+|x|} \lt 1.$
$\therefore |\frac {x}{1+|x|}| \lt 1 \Rightarrow -1 \lt \frac {x}{1+|x|} \lt 1$.
