Uniqueness of total derivative on non-open sets Let $A\subseteq \mathbf{R}^2$ and $ f : A \to\mathbf{R}^m $ is differentiable at $c \in A\cap A'$ (where $A'$ is the set of all limit points of $A$), also there exist two linearly independent vectors $a, b \in \mathbf{R}^2$ and $\delta > 0$ such that $c + ta$ and $c + tb$ are $\in A$ for $t \in [0,\delta)\text{ or }(-\delta, 0].$ Prove that Total derivative is then unique.
For total derivative I mean a linear map $L \in \mathscr{L}(\mathbf{R}^2, \mathbf{R}^m) $ such that $$\lim_{x \to c} \frac {\Vert f(x) - f(c) - L(x-c)\Vert}{\Vert x - c\Vert} = 0$$
Honestly have no idea how to approach this
 A: I think I actually solved it.
Suppose there are 2 such linear maps $L_1$ and $L_2$, let $\epsilon > 0$ be fixed and then from differentiability, $\exists\delta_0>0$ such that $\forall x\in A$, $$d(x,c)<\delta_0 \implies d(H_1,(0,0)) < \frac{\epsilon}{2\cdot\Vert a\Vert},$$ where $$
H_1 = \frac{\Vert f(x)-f(c)-L_1(x-c)\Vert}{\Vert x-c\Vert}.
$$ 
We define $H_2$ for $L_2$ analogously: now let $t \in (0,\delta) \cap (0,{\delta_0}/{\Vert a\Vert})$ and consider $x = c +ta \in A$. By our assumptions, 
$$
t = \frac{\Vert x-c\Vert}{\Vert a\Vert}\:\text{ and }\:a = \frac{\Vert a\Vert}{\Vert x-c\Vert}(x-c),
$$
thus
$$
\begin{split}
\Vert L_1(a) - L_2(a)\Vert &=\Vert a\Vert \cdot\left\Vert\frac{L_1(x-c)-L_2(x-c)}{\Vert x-c\Vert}\right\Vert\\
&=\Vert a\Vert \cdot\left\Vert \frac{-f(x)+f(c)+L_1(x-c)}{\Vert x-c\Vert }+\frac{f(x)-f(c)+L_2(x-c)}{\Vert x-c\Vert}\right\Vert\\
&\le
\Vert a\Vert \left[\frac{\Vert f(x)-f(c)-L_1(x-c)\Vert}{\Vert x-c\Vert}+\frac{\Vert f(x)-f(c)-L_2(x-c)\Vert}{\Vert x-c\Vert}\right]\\
&\lt\Vert a\Vert\cdot\left[\frac{\epsilon}{2\cdot\Vert a\Vert}+\frac{\epsilon}{2\cdot\Vert a\Vert}\right]=\epsilon,
\end{split}
$$
therefore $L_1(a) = L_2(a)$. Repeating the whole proof again for the vector $b$ we get $L_1(b) = L_2(b)$, and since $\{a,b\}$ is a basis for $\mathbf{R}^2$ we get $L_1 = L_2$.
