A function $f:U\subset\mathbb{R}^n \rightarrow \mathbb{R}^m$, $U$ open, is differentiable in $p \in U$ if there exists a linear transformation $T:\mathbb{R}^n \rightarrow \mathbb{R}^m$ such that $f(p+v)=f(p)+T(v)+R(v)$, where $R(v)$ satisfies $lim _{v\rightarrow 0}\dfrac{R(v)}{|v|}=0$, for all $v\in\mathbb{R}^n$ with $p+v\in U$.
That said, if $f$ as above is differentiable and f'(x)=T, $\forall x\in\mathbb{R}^n$. I need to show that there is an $a\in\mathbb{R}^n$ such that $f(x)=Tx+a$.
The problem I'm having is, how do I show that $R(v)=0$ for all $v$?