Can the directional derivative be defined as $\lim_{\|v\|\rightarrow 0}\frac{f(\mathbf{x} + \mathbf{v})-f(\mathbf{x})}{||\mathbf{v}||}$? I am thinking about the directional derivative. I think that the easiest way how to express it is
\begin{equation}
\frac{\partial f(\mathbf{x})}{\partial \mathbf{v}} = \lim_{||v|| \rightarrow 0} \frac{f(\mathbf{x} + \mathbf{v})-f(\mathbf{x})}{||\mathbf{v}||},
\end{equation}
but the directional derivative is usually defined as
\begin{equation}
\frac{\partial f(\mathbf{x})}{\partial  \mathbf{v}} = \lim_{h \rightarrow 0} \frac{f(\mathbf{x} + h \mathbf{v})-f(\mathbf{x})}{h}.
\end{equation}
Can you rigorously explain the transition from first and second definition?
Thanks!

Edit:
Just to make it clear.
The first definition is primarily wrong because the orientation of the directional vector is not fixed. So if I correct it like this (switching to conventional notation)
\begin{equation}
\nabla_{\mathbf{v}} f(\mathbf{x}) = \lim_{||\mathbf{v}|| \to 0} \frac{f(\mathbf{x} + ||\mathbf{v}||\mathbf{\hat{v}})-f(\mathbf{x})}{||\mathbf{v}||},
\end{equation}
it makes a little bit more sense ($ \mathbf{\hat{v}}$ denotes unit vector). BUT the norm allows to get close to zero just from right side ($ ||\mathbf{v}|| \to 0+ $) and the limit makes sense even for vector reversal (~ negative norm). So we can actually use any scalar $h$ scaling the vector and write the derivative as
\begin{equation}
\nabla_{\mathbf{v}} f(\mathbf{x}) = \lim_{h \to 0} \frac{f(\mathbf{x} + h\mathbf{\hat{v}})-f(\mathbf{x})}{h}
\end{equation}
or you can find it equivalently written as
\begin{equation}
\nabla_{\mathbf{v}} f(\mathbf{x}) = \lim_{h \to 0} \frac{f(\mathbf{x} + h\mathbf{v})-f(\mathbf{x})}{h||\mathbf{v}||}.
\end{equation}
 A: Your first definition makes no sense really, since you're using $\Delta \mathbf{x}$ simultaneously as a parameter and variable (which you're taking a limit over). The second (correct) definition thus uses a separate variable over which the limit is taken, used to scale the positional vector.
A: Your definition is not "directional," since $v$ in your definition can be in any direction. Note, if you were just in $1$ dimension, the limit $$\lim_{\delta\to 0} \frac{f(x+\delta)-f(x)}{|\delta|}$$ isn't even defined when $f$ is normally differentiable, because when $\delta\to 0-$ the limit is $-f'(x)$and when $\delta\to 0+$ the limit is $f'(x)$.
Essentially, the "directional derivative" can be seen as taking $\mathbf x$ and $\mathbf v$ and defining a new function on the real numbers: $g(h) = f(\mathbf x + h\mathbf v)$. This $g$ computes the value of $f$ on the line through $\mathbf x$ in the direction of $\mathbf v$, and the directional derivative of $f$ at $\mathbf x$ in the direction $\mathbf v$ is defined as $g'(0)$.
