Epsilon-Function Differentiability Condition

Theorem
Let $\mathbb K$ be either $\R$ or $\C$.

Let $f: D \to \mathbb K$ be a continuous mapping, where $D \subseteq \mathbb K$ is an open set.

Let $z \in \mathbb K$.

Then $f$ is differentiable at $z$ iff there exist $\alpha \in \mathbb K$ and $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:


 * $f \left({z + h}\right) = f \left({z}\right) + h \left({\alpha + \epsilon \left({h}\right) }\right)$

where $B_r \left({0}\right)$ denotes an open ball of $0$, and $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \mathbb K$ is a continuous mapping with $\displaystyle \lim_{h \to 0} \ \epsilon \left({h}\right) = 0$.

If the conditions are true, then $\alpha = f' \left({z}\right)$.

Also see

 * Characterization of Differentiability