I want to show that $x\neq O(x^2)$ for $x\to0$, using contradiction. So assume $x=O(x^2)$. That means there exists $C\geq0$ and $d>0$ such that:
$$
\vert x\vert\leq Cx^2\quad\text{for all }x\in(-d,d).
$$
No I would like to divide by $x$, so I have to consider 2 cases: $x\in(-d,d)\setminus\{0\}$ and $x=0$. However, I wouldn't know how to divide an absolute value by $x$, without making stuff really complicated, by again considering cases. Is there an easy (easier?) way to do this?
Definition
Let $f$ be a map $A \to B$. Retraction for map $f$ is a map $r$ such that $r \circ f = I_{A}$.
Let $f$ be a map $A \to B$, $h$ a map from $A \to C$. Then the problem of finding a map $g$ $B \to C$ such that $g \circ f = h$ is called a determination problem.
Let $f$ be map $A \to B$,...
