Availability of derivative for multivariable functions: Are these conditions equivalent? In Hubbard's multivariable calculus book a function $f:\mathbb R^m \mapsto \mathbb R^n$ has a derivative at a point $a \in \mathbb R^m$ if the following equation holds:
$$\lim_{\vec h \to \vec 0}\frac{1}{||\vec h||}\left( (f(a+\vec h)-f(a))-[Df(a)]\vec h) \right)=\vec 0$$
But in one example this equation is used without explanation:
$$\lim_{\vec h \to \vec 0}\frac{1}{||\vec h||}|| (f(a+\vec h)-f(a))-[Df(a)]\vec h) ||= 0$$ 
They seem to represent exactly the same thing but I don't want to come to a conclusion on my own. Are these conditions equivalent? 
 A: Well this directly follows from the definition of limits in normed vector spaces. Let's start with a simple example: Let $X$ a normed vector space with norm $\vert\vert\cdot \vert\vert$ and $(x_n)_{n\geq0}, x \in X$ then: 
$$ x_n \to x, \; n\to \infty  $$
$$\Leftrightarrow x_n - x\to 0, \; n\to \infty  $$
$$\Leftrightarrow \forall \varepsilon>0 \; \exists \; N\geq 0: \vert\vert x_n - x \vert \vert < \varepsilon \; \forall n\geq N $$
$$\Leftrightarrow \vert\vert x_n - x \vert \vert \to 0, \; n \to \infty $$
Notice that the final limit is a limit in $\mathbb R$. Similarly we can always prove the limit of a function $g: \mathbb R^m \to \mathbb R^n$ in terms of the limit of a function $\widetilde{g}: \mathbb R^m \to \mathbb R$. In your case:
$$\lim_{\vec h \to \vec 0}\frac{1}{||\vec h||}\left( (f(a+\vec h)-f(a))-[Df(a)]\vec h) \right)=\vec 0$$
$$\Leftrightarrow \forall \varepsilon>0 \; \exists \; \delta >0: \big\vert\big\vert\frac{1}{||\vec h||}\left( (f(a+\vec h)-f(a))-[Df(a)]\vec h) \right)\big\vert\big\vert <\varepsilon \; \forall \vec h \text{ with }\vert\vert \vec h \vert\vert < \delta  $$
$$ \Leftrightarrow \lim_{\vec h \to \vec 0}\big\vert\big\vert\frac{1}{||\vec h||}\left( (f(a+\vec h)-f(a))-[Df(a)]\vec h) \right)\big\vert\big\vert=0$$
