Showing uniqueness of Fréchet Derivative (gradient) Given the following definition of differentiability, I am trying to understand the uniqueness of the gradient proof.
Let $\bf{E}$ be a finite dimensional vector space over $\bf{R}$. $f:\bf{E}\to\bf{R}$ is differentiable at $\bf{x}\in E$ if there exists $\bf{g} \in {E}^{*}$ (the dual space, which my text identifies with $\bf{E}$) such that: $$\lim_{\bf{h}\to0}\frac{f(\textbf{x+h})-f(\textbf{x})-\langle \textbf{g,h}\rangle}{||\bf{h}||}=0$$ Such a vector $\bf{g}$ is denoted the gradient of $f$ at $\bf{x}$.
The author gives this approach: Suppose two vectors $\bf{g_1,g_2}$ satisfy this equation. Then subtracting the limits yields $$\lim_{\bf{h\to0}}\frac{\langle \bf{g_1-g_2},h\rangle}{||\bf{h}||}=0$$
They claim that this immediately shows that $\bf{g_1 = g_2}$. I do not understand this last step. Any insight would be appreciated.

Hint: Suppose $g_1-g_2\neq 0$ and take $h$ to have the same direction of $g_1-g_2,$ then $h=\frac{||h||}{||g_1-g_2||}(g_1-g_2)$ and make $||h||$ go to zero, where does the limit goes?
