show that $\textbf{u} = \textbf{v}$ if $\langle \textbf{u}, \textbf{h}\rangle + o(\|h\|) = \langle \textbf{v}, \textbf{h}\rangle + o(\|h\|)$ Let $\textbf{u}$ and $\textbf{v}$ be two vectors in $\mathbb{R}^n$ such that:
$$
\langle \textbf{u}, \textbf{h}\rangle + o(\|h\|) = \langle \textbf{v}, \textbf{h}\rangle + o(\|h\|) \qquad (\textbf{h}\in \mathbb{R}^n)
$$
show that $\textbf{u} = \textbf{v}$
$$
\exists \varepsilon_1, \varepsilon_2; \qquad \varepsilon_1(h)\xrightarrow[h \to 0]{}0 \text{ and } \varepsilon_2(h)\xrightarrow[h \to 0]{}0
$$
$$
\langle \textbf{u}, \textbf{h} \rangle + \varepsilon_1 (\textbf{h})\|\textbf{h}\| = \langle \textbf{v}, \textbf{h} \rangle + \varepsilon_2 (\textbf{h})\|\textbf{h}\|
$$
$$
\implies \langle \textbf{u} - \textbf{v}, \textbf{h}\rangle = \|\textbf{h}\|\underbrace{(\varepsilon_2(\textbf{h}) - \varepsilon_1(\textbf{h}))}_{\varepsilon(h)\xrightarrow[h \to 0]{}0}
$$
$$
\langle \textbf{u}- \textbf{v}, \textbf{h}\rangle = o(\|\textbf{h}\|)
$$
Suppose that $\textbf{u} \neq \textbf{v} \implies \|\textbf{u} - \textbf{v}\| \neq 0$, we can take $\textbf{h} = \frac{\textbf{u} - \textbf{v}}{\|\textbf{u}- \textbf{v}\|}$. Then $\langle \textbf{u}- \textbf{v}, \frac{\textbf{u}- \textbf{v}}{\|\textbf{u}- \textbf{v}\|}\rangle = \|\textbf{u}- \textbf{v}\| = o(\|\textbf{u}- \textbf{v}\|) \implies$ Contradiction. Therefore, if $\forall \textbf{h}\in \mathbb{R}^d \langle \textbf{u}, \textbf{v} \rangle + o(\|\textbf{h}\|) = \langle \textbf{v}, \textbf{v} \rangle + o(\|\textbf{h}\|) \textbf{u} = \textbf{v}$
My question: 
Why is  $\langle \textbf{u}- \textbf{v}, \frac{\textbf{u}- \textbf{v}}{\|\textbf{u}- \textbf{v}\|}\rangle = \|\textbf{u}- \textbf{v}\| = o(\|\textbf{u}- \textbf{v}\|)$ a Contradiction?
 A: Start from the assumption, namely that for all $\textbf{h}\in\mathbb R^n$,  $$\langle \textbf{u}, \textbf{h}\rangle + o(\|\textbf h\|) = \langle \textbf{v}, \textbf{h}\rangle + o(\|\textbf h\|) $$
Subtract the right-hand side from the left-hand side:
$$\langle \textbf{u-v}, \textbf{h}\rangle = o(\|\textbf h\|)\tag{1}$$
That's because $o(\|\textbf h\|)-o(\|\textbf h\|)=o(\|\textbf h\|)$.
Now, Suppose $\textbf{u}\neq \textbf{v}$. Let $t\in \mathbb R$, and define $\textbf{h}=t(\textbf{u}-\textbf{v})$. Plug this into  $(1)$, you get, for all  $t\in\mathbb R$:
$$\begin{split}
t\|\textbf{u}-\textbf{v}\|^2 &=o(|t|\|\textbf u-\textbf v\|)\\
&=o(t)\cdot\|\textbf{u}-\textbf{v}\|\\
&=o(1)\cdot |t| \cdot \|\textbf u-\textbf v\|
\end{split}$$
which simplifies  into
$$\|\textbf{u}-\textbf{v}\|=o(1)$$
Consequently, by letting $t\rightarrow 0$, $\|\textbf{u}-\textbf{v}\|=0$.
A: If $u \ne v$ then there exists a nonzero vector $h$ such that $\langle u-v,h\rangle \ne 0$.
However, considering $th$ for $t > 0$ we have $$0\ne \langle u-v,h\rangle = \frac1t\langle u-v,th\rangle = \|h\|(\varepsilon_1(th) - \varepsilon_2(th)) \xrightarrow{t\to 0} 0$$
which is a contradiction.
