Let $\pmb F : U \subseteq \mathbb{R}^n \to \mathbb{R}^m$ is differentiable at $\pmb a \in U$ and let $\pmb a $ is an cluster point of $U$ if there exists a linear map $T : \mathbb{R}^n \to \mathbb{R}^m $ satisfying: $$\big(\forall \epsilon > 0 \in \mathbb{R} \big) \big(\exists \delta_{(\epsilon,\pmb a)}>0 \in \mathbb{R}\big) $$ $$\text{such that} :$$ $$\parallel \big(\pmb x-\pmb a \big) \parallel \leq \delta \ \ \text{for every } \ \ \pmb x \in \mathbb{R}^n \Longrightarrow \parallel \pmb F(\pmb x)- \pmb F(\pmb a) - T(\pmb x -\pmb a)\parallel \leq \epsilon \parallel \pmb x -\pmb a\parallel$$
The map $T$ is unique, denoted $df_{\pmb a} $ and called ‘differential (or ‘derivative’) of $f$ at $\pmb a$ . Another notation for this is: $$\pmb F(\pmb x)= \pmb F(\pmb a) + df_{\pmb a}(\pmb x -\pmb a)+\pmb r(\pmb x)$$
I have two questions :
$1-)$ How prove that $T$ is unique ?
$2-)$ What is function of $\pmb r$ ? What is domain and range of function $\pmb r$ ?
Let $T'$ be another differential, $\|x-a\|\leq \delta_n$ implies that $\|F(x)-F(a)-T(x-a)\|\leq 1/n\|x-a\|$ and $\|F(x)-F(a)-T'(x-a)\|\leq 1/n\|x-a\|$.
We deduce that $\|T(x-a)-T'(x-a)\|\leq \|(T(x-a)-(F(x)-F(a))+(F(x)-F(a)-T'(x-a)\|\leq \|F(x)-F(a)-T(x-a)\|+\|F(x)-F(a)-T'(x-a)\|\leq 2/n \|x-a\|$.
Let $b$ such that $\|x-b\|=1$, we have $\|\delta_n(x-b)\|=\delta_n$ impliest that $\|T(\delta_n(x-b))-T'(\delta_n(x-b))\|\leq {2\over n}\|\delta_n(x-b)\|$. This implies that $\|(T-T')(x-b)\|\leq {2\over n}$ for every $n>0$, we deduce that $(T-T')(x-b)=0$ and $T-T'=0$. Since for every $a\neq x$, $(T-T')({{x-a}\over{\|x-a\|}})=0$.
By definition, $r(x)= F(x)-F(a)-df_a(x-a)$.