Again I'm having a hard time understanding the notation of a proof concerning the statement that if $U \subset \mathbb{R}^n$ is open, $\vec f:U \rightarrow \mathbb{R}^m$ is a mapping which is differentiable in $\vec x_0 \in U$, then it is differentiable in direction of every vector $\vec v \in \mathbb{R}^n$ with:
$\partial \vec v \vec f (\vec x) = d \vec f (\vec x_0) \vec v$.
So the proof is given by:
$\vec f (\vec x_0 +h \vec v) = \vec f(\vec x_0)+h \cdot d \vec f(\vec x_0)\vec v + \vec \gamma (h \vec v)$ with:
$lim_{h \rightarrow 0} \frac{||\gamma (h \vec v)||}{||h \vec v||} = lim_{h \rightarrow 0} \frac{||\gamma (h \vec v)||}{|h|} \cdot \frac{1}{||\vec v||} = 0$
therefore exists:
$lim_{h \rightarrow 0} \frac{\vec f (\vec x_0 +h \vec v) - \vec (\vec x_0)}{h} = d \vec f(\vec x_0) \vec v =: \partial_{\vec v} \vec f (\vec x_0)$.
The first question I have is, how can I understand $\vec v$? Is it some random vector in some random direction or does it have any properties that I'm missing? Also, since we defined the differential of a function in mulitivariate space earlier by:
$A \vec h = d \vec f(\vec x_0) \vec v$ if an $A : \mathbb{R}^n \rightarrow \mathbb{R}^m$ exists with porperties:
$\vec f (\vec x_0 + \vec h) - \vec (\vec x_0) - A \vec h = \vec \gamma (\vec h)$ while $\vec \gamma ( \vec h) = 0(||\vec h||) \Leftrightarrow \lim_{h \rightarrow 0} \frac{||\gamma (h \vec v)||_{\mathbb{R}^m}}{||h||_{\mathbb{R}^n}} = 0$
how can I think of the product $h \vec v$? And why does $h$ just disappear in the formula that the proof arrives at?
Thanks for any clarification on the proof and on the notation, I'd also appreciate any sources that are written in similar notation since all I can find on the internet is written using different notation.