Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be a function which partial derivatives $\frac{\partial f}{\partial x_i}$ are defined at $\vec{a}\in \mathbb{R}^n$ for $1\leq i \leq n$, and let $\vec{u}$ be a unit vector. Here are two definitions of the directional derivative $f_{\vec{u}}(\vec{a})$:
$$\lim_ {h\rightarrow 0}\frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h}$$
$$\nabla f(\vec{a})\cdot \vec{u}$$
I'm seeking to prove that the two are equivalent.
There are two proofs of the above here, but if the proofs are valid then I do not understand how. In particular:
The second proof, written by me, applies the Mean Value Theorem on the partial derivatives of $f$, thus it assumes that the partial derivatives are continuous on an open ball around $\vec{a}$.
The first proof by user137731 makes use of the fact that
$$f(\vec{a}+\vec{h})=f(\vec{a})+\nabla f(a)\cdot \vec{h} + o(\vec{h})$$
where $o(\vec{h})/h\rightarrow 0$ as $\vec{h}\rightarrow 0$. The user derives the equation by means of Taylor's theorem. If Taylor's theorem can be applied here without presupposing the continuiuty of the partial derivatives or of the function $f$, then someone please point me how. The equation can also be derived easily by assuming that the total differential exists at $\vec{a}$ (I'm even wondering if the equation being true implies the existence of the total derivative), yet, again, we are not allowed to assume that the function is differentiable at $\vec{a}$.
