I have this function:
$$f(x)=\frac{x}{x^2+1}$$
it's clear that the domain is all real numbers, because the denominator never become zero. I've done the following procedure to found the range of $f(x)$:
$\operatorname{Rec} f=\left \{{y\in \mathbb{R}:\exists x\in \operatorname{Dom} f,y=f(x)} \right \} \\ \operatorname{Rec} f=\left \{{y\in \mathbb{R}:\exists x\in \mathbb{R},y=\dfrac{x}{x^2+1}} \right \} \\ \operatorname{Rec} f=\left \{{y\in \mathbb{R}:\exists x\in \mathbb{R},x=\dfrac{1\pm \sqrt{1-4y}}{2y}} \right \} \\ \operatorname{Rec} f=\left \{{y\in \mathbb{R}:\left (1-4y^2 \right )\geq 0\, \wedge\, y\neq 0 } \right \}\\ \operatorname{Rec} f=[-\tfrac{1}{2},\tfrac{1}{2}]-\left \{ 0 \right \}$
But, the function is zero when $x=0$ $(f(0)=0)$, and the range of the function is $[-\tfrac{1}{2}, \tfrac{1}{2}]$. So I don't know where is my error in the procedure.