I was trying to understand why $0<\frac{x}{1-x}<\infty$ became $0I was trying to understand why $0<\frac{x}{1-x}<\infty$, became $0<x<1$.
I was thinking of multiplying the inequalities to $1-x$, but $(1-x)\infty$ won't yield to $1$, right?
 A: Let $f(x) = \frac{x}{1-x}$
If $x = 1$, the function is not well defined.
If $x > 1$, $x>0$ but $1-x<0$, hence $f(x) <0$.
If $x \leq 0$, $x \leq 0$, and $1-x > 0$, hence $f(x) \leq 0$.
Hence the only possibility left for $f(x) >0$ would be $0<x<1.$
Alternatively, first note that $x \neq 1$. After which for $0< \frac{x}{1-x}$, we multiply both sides by $(1-x)^2$.
Hence $$0<x(1-x)$$
Solving the inequality gives us $0<x<1.$
Be careful that we can't just multiply $1-x$ directly without first checking if you might have multiplied with a negative number.
A: I would consider the $< \infty$ redundant. You're looking for where the function $\frac{x}{1 - x}$ is positive. I would solve it by noticing that you have a quotient of two quantities $x$ and $1 - x$. In order to be positive, we need both of them to be positive or both to be negative.
We have $x > 0$ and $1 - x > 0$, if and only if $0 < x < 1$. On the other hand, $x < 0$ and $1 - x < 0$ if and only if $x < 0$ and $x > 1$, which is impossible. Thus, the solution is $0 < x < 1$.
