Range of a 1-2 function $$f(x)=\frac x {x^2+1}$$ I want to find range of $f(x)$ and I do like below . If someone has different Idea please Hint me .
Thanks in advanced .
This is 1-1 function $\\f(x)=\dfrac{ax+b}{cx+d}\\$,
This is 2-2 function $\\f(x)=\dfrac{ax^2+bx+c}{a'x^2+b'x+c'}\\$,
This is 1-2 function $\\f(x)=\dfrac{ax+b}{a'x^2+b'x+c'}\\$
 A: Method 1
$$y=\dfrac{x}{x^2+1} \to yx^2-x+y=0  \\\Delta \geq 0  \to 1-4(y)(y) \geq0  \\y^2 \leq \dfrac14 \\ y \in \left[-\dfrac12,\dfrac12\right]$$
Method 2
It is well known that$$\left|x+\frac1x\right| \geq 2 \to y=\dfrac{x}{x^2+1} \to \dfrac1y=\dfrac{x^2+1}{x}=x+\dfrac1x \\\to |\dfrac1y|\geq 2 \\|y| \leq \dfrac12 \space y \in\left[-\dfrac12,\dfrac12\right]$$
Method 3
$$y=\frac{x}{x^2+1} \to y'=\frac{1-x^2}{(x^2+1)^2}=0 \to x=\pm1 \\ \begin{cases}x=1  & f(1)=\frac12 \\x=-1  & f(1)=-\dfrac12\\
 x\to -\infty & \lim_{x \rightarrow -\infty}f(x) \to 0\\x\to -\infty & \lim_{x \rightarrow +\infty}f(x) \to 0\end{cases} \to y \in\left[-\frac12, \dfrac12 \right] $$
Method 4
By substituting $x=\tan \alpha$ we have 
$$y=\dfrac x {x^2+1}=y=\frac{\tan \alpha}{\tan^2 \alpha+1}=\frac{\dfrac{\sin \alpha}{\cos \alpha}}{\left(\dfrac{\sin \alpha}{\cos \alpha}\right)^2+1}=\\=\sin \alpha \cdot \cos  \alpha=\frac22 \sin \alpha \cdot\cos  \alpha =\frac12 \sin 2\alpha \\ -1 \leq \sin 2\alpha \leq 1 \\ \to -\frac12 \leq \frac12 \sin 2\alpha \leq \frac12 \to y \in \left[-\dfrac12,\dfrac12\right]$$
A: Method 5
$0\le(x+1)^2 = (x^2+1)+2x$ dividing by $(x^2+1)$ we have $0\le1+2y$
$0\le(x-1)^2 = (x^2+1)-2x$ dividing by $(x^2+1)$ we have $0\le1-2y$
And $y\in[-\frac12,\frac12]$.
