Definitions of differentiability I have seen two definitions of differentiability of a real valued function and I wonder why they are equivalent.
The first definition: For a function $\mathbf{F}:\mathbb R^n\to \mathbb R$ is differentiable at $\mathbf x$, if there exist a matrix $\mathbf{DF}(\mathbf x)$ such that $$\lim_{\Delta\mathbf x\to\mathbf0}\frac{\mathbf F(\mathbf x+\Delta\mathbf x)-\mathbf F(\mathbf x)-\mathbf{DF}(\mathbf x)\Delta\mathbf x}{||\Delta\mathbf x||}=0$$
It turns out that $\mathbf{DF}=\nabla \mathbf F$
So if $\mathbf F$ is differentiable, $$\lim_{\Delta\mathbf x\to\mathbf0}\frac{\Delta\mathbf F-\nabla \mathbf F\cdot\Delta\mathbf x}{\Delta \mathbf x}=0$$
The second definition: If $\mathbf F$ is differentiable, 
$$\Delta\mathbf F=\nabla\mathbf F\cdot\Delta\mathbf x+\epsilon\cdot\Delta\mathbf x$$ 
where $\epsilon=(\epsilon_1,\epsilon_2,...,\epsilon_n)$ and $\epsilon\to\mathbf0$ as $\Delta \mathbf x\to \mathbf0$
I know how to prove 2 implies 1 by Cauchy Schwarz inequality ($\epsilon\cdot\Delta\mathbf x\le||\epsilon||||\Delta\mathbf x||$) but I don't know how to prove 1 implies 2
 A: For the other direction, define a function $k:\mathbb R^n\setminus\{\mathbf0\}\rightarrow\mathbb R$ by
$$
k=k(\Delta\mathbf x)=\frac{\Delta\mathbf F-\nabla \mathbf F\cdot\Delta\mathbf x}{||\Delta \mathbf x||}
$$
then $\Delta\mathbf F-\nabla \mathbf F\cdot\Delta\mathbf x=k\cdot||\Delta\mathbf x||$. Since $\Delta\mathbf x\neq\mathbf 0$ we have $||\Delta\mathbf x||\neq 0$. Thus we may define 
$$
\mathbf\epsilon=k\cdot\frac{\Delta\mathbf x}{||\Delta\mathbf x||}
$$
in order to have $\mathbf\epsilon\cdot\Delta\mathbf x=k\cdot||\Delta\mathbf x||=\Delta\mathbf F-\nabla \mathbf F\cdot\Delta\mathbf x$. Finally, since we have assumed 1 to hold we know that $k\longrightarrow 0$ for $\Delta\mathbf x\longrightarrow \mathbf0$ and since the normed vector $\dfrac{\Delta\mathbf x}{||\Delta\mathbf x||}$ is bound to be on the unit sphere having length $1$ it follows that $\mathbf\epsilon=k\cdot\dfrac{\Delta\mathbf x}{||\Delta\mathbf x||}\longrightarrow\mathbf 0$. This proves that $\mathbf{(1)}\implies\mathbf{(2)}$.
