Proving the uniqueness derivative of vector-valued function For a vector-valued function $f:c\in E\subseteq \Bbb R^n\to\Bbb R^m$,  I want to prove that the linear transformation $T$ such that $\displaystyle\lim_{h\to 0}\frac{f(c+h)-f(c)-T(h)}{|h|}=0$ is unique.
My attempt:
Suppose $T_1,T_2:\Bbb R^n\to\Bbb R^m$ such that $\displaystyle\lim_{h\to 0}\frac{f(c+h)-f(c)-T_1(h)}{|h|}=0$ and $\displaystyle\lim_{h\to 0}\frac{f(c+h)-f(c)-T_2(h)}{|h|}=0$. Then $\displaystyle\lim_{h\to 0}\frac{T_2(h)-T_1(h)}{|h|}=0$. Suppose  $T_1\neq T_2$, then there exists $v\neq 0$ such that $T_1(v)\neq T_2(v)$. Compositing into the function $t\mapsto tv$. Then we can get $\displaystyle\lim_{t\to 0}\frac{T_2(tv)-T_1(tv)}{|tv|}=0$. However, how can I deduce the contradition?
 A: Hint
Note that 
$$\lim_{t\to 0^+}\frac{T_2(tv)-T_1(tv)}{|tv|}=\lim_{t\to 0^+}\frac{t(T_2(v)-T_1(v))}{|tv|}=\lim_{t\to 0^+}\frac{T_2(v)-T_1(v)}{|v|}.$$
A: Suppose $T$ and $S$ are two correct derivatives.  Write $U = T-S$.  Then
$$
|U\mathbf{h}| \leq  |f(\mathbf{c}+\mathbf{h})-f(\mathbf{c})-T\mathbf{h}| + |f(\mathbf{c}+\mathbf{h})-f(\mathbf{c})-S\mathbf{h}|
$$
should convince you that $\lim_{\mathbf{h} \to \mathbf{0}} \frac{|U\mathbf{h}|}{|\mathbf{h}|} = 0$.  Fixing $\mathbf{h} \neq \mathbf{0}$ for the moment, 
$$
\lim_{t \to 0} \frac{|U(t\mathbf{h})|}{|t\mathbf{h}|} = 0.
$$
But the latter has nothing to do with $t$ by linearity, so it must be that $U\mathbf{h} = \mathbf{0}$ for all $\mathbf{h}$.  Hence $U=0$ and $S=T$.
A: Let$T_1(v)\neq T_2(v)$ for a fixed $v\in \mathbb{R}^n$. Then
$$
T_1\left(\frac{1}{\|v\|}v\right)- T_2\left(\frac{1}{\|v\|}v\right)\neq 0
$$ We know that
$$
\lim_{t\to 0}\frac{f(c+tv)-f(c)-T_1(tv)  }{\left\|tv \right\|}=0
\\
\lim_{t\to 0}\frac{f(c+tv)-f(c)-T_2(tv)  }{\left\|tv \right\|}=0
$$
implies 
$$
\lim_{t\to 0}\frac{T_1(tv)-T_2(tv)}{\left\|tv \right\|}=0.
$$
But we have 
$$
\lim_{t\to 0}\frac{T_1(tv)-T_2(tv)}{\left\|tv \right\|}
=
\underbrace{
\left(
T_1\left(\frac{1}{\|v\|}v\right)- T_2\left(\frac{1}{\|v\|}v\right) 
\right)
}_{\rm constant}
\lim_{t\to 0}\frac{t}{|t|}
$$
and we know that $\frac{t}{|t|}=-1$ or $\frac{t}{|t|}=+1$. This implies $\lim_{t\to 0}\frac{t}{|t|}=\pm1$. Then
$$
\lim_{t\to 0}\frac{T_1(tv)-T_2(tv)}{\left\|tv \right\|}
=\pm \left(
T_1\left(\frac{1}{\|v\|}v\right)- T_2\left(\frac{1}{\|v\|}v\right) 
\right)
\neq 0.
$$
