How to prove the Bijection of an Interval $(-1,1)$ to $\mathbb{R}$? How can I prove that this image is bijective?
$$f: (-1,1) \longrightarrow \mathbb{R}, \quad  x \longmapsto \frac{x}{1-x^2}
$$
is bijective without the use of the Steepness of the slope?
 A: This is a sketch of how; filling out details is left as an exercise.
$y=\frac{x}{1-x^2}$ implies $yx^2+x-y=0$, which has roots of product $-1$ so only one of them can be in $(-1,\,1)$. This proves $f$ never takes the same value twice.
A: If $0<x<1,$ then as the numerator $x$ increases, the positive denominator $1-x^2$ decreases, so the quotient as a whole grows. And the denominator approaches $0$ as the numerator approaches $1,$ so this goes up to $+\infty.$
If strictly increases then it's one-to-one, and if it's continuous and goes from $0$ up to $+\infty$ as as as its argument goes from $0$ to $1,$ then it's onto the interval $(0,+\infty).$
But the intermediate value theorem is needed for that.
You can do something similar for $-1<x<0,$ and the case $x=0$ is easy to understand, and the value of the function is negative for $-1<x<0$ and positive for $0<x<1,$ so that doesn't upset one-to-one-ness.
You can also do this with algebra, by solving a quadratic equation, provided you believe every nonnegative number has a nonnegative square root.
A: Because $f(x)=\frac{x}{1-x^2}$ is a continuous function on $(-1,1)$,  $$\frac{x}{1-x^2}-\frac{y}{1-y^2}=\frac{(x-y)(1+xy)}{(1-x^2)(1-y^2)},$$
$$\lim_{x\rightarrow-1^+}\frac{x}{1-x^2}=-\infty$$ and
$$\lim_{x\rightarrow1^-}\frac{x}{1-x^2}=+\infty$$ 
