Conditions for regarding only the numerator if a fraction equals zero This is closely related to this question, see also the follow-up question.
Usually we say that if we want to find solutions to $\frac{f(x)}{g(x)} = 0$, it is enough to check where $f(x) = 0$ and $g(x) \neq 0$.
However, consider for example
$$\frac{x^2}{1 + \frac{1}{1-x}} = 0.$$
The numerator gives us the solution $x = 0$ only, but rewriting as 
$$\frac{x^2 - x^3}{2 - x}=0$$
reveals that $x = 1$ is a solution as well. 
Alternatively, we could think about the denominator going to infinity and therefore the whole term to $0$:
$$\lim\limits_{x \rightarrow 1} \frac{x^2}{1 + \frac{1}{1-x}} = 0.$$
What are the conditions for considering the numerator only?
 A: Your question can't be answered with a rule about "when to consider only the numerator ..."
The expression
$$ \frac{x^2}{1 + \frac{1}{1-x}} 
$$
is not defined when $x=1$ or $x=2$, so it has no value at those points. The expression 
$$
\frac{x^2 - x^3}{2 - x}
$$
is not defined when $x = 2$.
The two expressions agree where they are both defined.
You can only look for values of $x$ that yield $0$ at places where an expression makes sense. That's $x=0$ for the first expression and $x=0,1$ for the second.
A: The way I was taught is that $x = 1$ is not a solution to 
$$
\frac{x^2}{1 + \frac{1}{1-x^2}} = 0.
$$
At some point in your factoring, you must have implicitly used the fact that $1 + \frac{1}{1 - x^2} \neq 0$. Even though $x=1$ looks like a solution, it isn't, because you had to divide by something that's $0$ when $x=1$.
Following these rules, you may assume that if $\frac{f(x)}{g(x)} = 0$, then $f(x) = 0$ and $g(x) \neq 0$, which is the kind of world I want to live in.
A: The solutions must belong to the domain of the function. $x=1$ does not.
You may not freely rewrite. For example, $\dfrac{x^2}x$ has no root, nor $\dfrac1{\frac1x}$.
The complete condition is $$x\in\text{dom}(f)\land x\in\text{dom}(g)\land g(x)\ne0\land f(x)=0.$$
