Definition of differentiability in higher dimensions I am taking vector calculus and I have been introduced to the following definition:
Let $A\subset \mathbb{R}^n$ and $f:A\to \mathbb{R}$. We say that $f$ is differentiable at a point $\boldsymbol{x}\in A$ if there is a linear map $T:\mathbb{R}^n\to \mathbb{R}$ such that:
$$\lim_{\boldsymbol{h}\to 0}\frac{|f(\boldsymbol{x}+\boldsymbol{h})-f(\boldsymbol{x})-T(\boldsymbol{h})|}{\|\boldsymbol{h}\|}=0$$
This definition makes sense to me, but I have not been able to find anywhere how to prove that this is well-defined meaning that the linear map $T$ is unique. Proof by contradiction seems logical but how can I prove this?
 A: It $T$ and $S$ both satisfy this property then $\frac {\|Th-Sh\| } {\|h\|} \to 0$. Fix any non-zero vector $u$ and take $h=cu$ where $c$ is a scalar to see that $Tu=Su$. Since $T$ and $S$ are linear maps and $Tu=Su$ for non-zero vectors  $u$ (and trivially for $u=0$) so $T=S$.
A: If there were two such $T$s, say $T_1$ and $T_2$, then you'd have
$$\lim_{h\to0}\frac{\|(T_1-T_2)(h)\|}{\|h\|}=0.$$
But that entails $T_1-T_2=0$, for if $(T_1-T_2)x\ne0$ then
$$\frac{\|(T_1-T_2)(\lambda x)\|}{\|\lambda x\|}\not\to0$$
as $\lambda\to0$.
A: Here is a more detailed treatment of the general case where the map is from $\mathbf{R}^n$ to $\mathbf{R}^m$.
Let $f:\mathbf{R}^n\rightarrow \mathbf{R}^m$ be a map that is differentiable at the point $a \in \mathbf{R}^n$. According to the definition, this means that there exists a linear map $T_a: \mathbf{R}^n\rightarrow \mathbf{R}^m$ such that $$\lim_{h\rightarrow 0} \dfrac{||f(a+h)-f(a)-T_a(h)||}{||h||}=0.$$
To show that $T_a$ is unique, suppose $S_a$ is any other such linear map. Let $b\in \mathbf{R}^n\setminus \{0\}$ (at the origin they both vanish as they are linear maps). Let us calculate $||T_a(b)-S_a(b)||$. Let $k \in \mathbf{R}\setminus\{0\}$, then we have the following: $$||T_a(b)-S_a(b)||=\frac{|k|||T_a(b)-S_a(b)||}{|k|}$$
By linearity of $T_a$, and $S_a$, we may rearrange the numerator as follows:
$$= \frac{||T_a(kb)-S_a(kb)||}{|k|}.$$
Now, we may add the vector $f(a+kb)-f(a)$ to the vector in the numerator and subtract it from the resulting vector to arrive at the following step:
$$=\frac{1}{|k|}||f(a+kb)-f(a)+f(a)-f(a+kb)+T_a(kb)-S_a(kb)||$$
Rearranging and using the triangle inequality yield
$$\leq \frac{||f(a+kb)-f(a)-S_a(kb)||}{|k|}+\frac{||f(a+kb)-f(a)-T_a(kb)||}{|k|}$$
To be able to use the definition, rewrite the expression as follows (multiply and divide by $||b||$):
$$=||b||\Bigg(\frac{||f(a+kb)-f(a)-S_a(kb)||}{|k|||b||}+\frac{||f(a+kb)-f(a)-T_a(kb)||}{|k|||b||}\Bigg)$$
Notice that the expression above goes to $0$ as $k\rightarrow 0$, thus $||T_a(b)-S_a(b)||=0$, which means that $T_a(b)=S_a(b)$ as desired. But $b\in \mathbf{R}^n\setminus \{0\}$ was arbitrary, thus $S_a=T_a$, and the given definition is well-defined.
