If $G_1$ and $G_2$ are affine transformations from affine spaces $E$ to $F$, and there exists a fixed $a\in E$ such that $\forall x\in E$
$$\lim_{\textbf{x}\to\ \textbf{a}} \frac{G_1(\textbf{x}) - G_2(\textbf{x})}{|\textbf{x} - \textbf{a}|} = \textbf{0},$$ prove that $G_1 = G_2$.
I'm not sure how to prove this. A proof by contradiction seems like the way to go, but I don't know how I would do that. Can someone suggest a way to prove this?
Since $G_1$ and $G_2$ are affine transformations, there exists linear transformations $\lambda_1$ and $\lambda_2$, such that $G_1(x)-G_1(a) = \lambda_1(x-a)$, $G_2(x)-G_2(a) = \lambda_2(x-a)$. Then $$\frac{G_1(x) - G_2(x)}{\|x-a\|} = (\lambda_1-\lambda_2)\frac{x-a}{\|x-a\|} + \frac{G_1(a) - G_2(a)}{\|x-a\|}.$$
If $\lambda_1\neq \lambda_2$, so we have a unit vector $\frac{x-a}{\|x-a\|}$ times a nonzero constant $(\lambda_1-\lambda_2)$ on right hand side. If $G_1(a) =G_2(a)$, $\frac{G_1(a) - G_2(a)}{\|x-a\|} = 0$, otherwise $\frac{G_1(a) - G_2(a)}{\|x-a\|}\rightarrow \infty$. Thus as $x\rightarrow a$, the norm
$$\|\frac{G_1(x) - G_2(x)}{\|x-a\|}\|\rightarrow\begin{cases} \|\lambda_1-\lambda_2\|,\ when\ G_1(a) =G_2(a)\\ \infty,\ when \ G_1(a) \neq G_2(a)\end{cases}.$$ In either case $\frac{G_1(x) - G_2(x)}{\|x-a\|}$ does not converge to $0$. Thus we must have $\lambda_1 = \lambda_2$ and $G_1(a ) = G_2(a)$. Now $G_1(x) = \lambda_1(x-a)+G_1(a) = \lambda_2(x-a)+G_2(a) = G_2(x),\forall x$. So $G_1=G_2$.