Let $V$ and $W$ be normed vector spaces, and $U \subset V$ be an open subset of $V$. A function $f: U \to W$ is called Frechet differentiable at $x \in U$ if there exists a bounded linear operator $A: V \to W$ s.t
$$ \lim_{ || h || \to 0 } \frac{||f(x+h) - f(x) - Ah||_W} {||h||_V}=0$$
Limit here is meant in the limit of function defined on metric space.., and the above expression as the function of arguements H in V. As a consequence, it must exist for all sequences $\langle h_n\rangle_{n=1}^{\infty}$ of non-zero elements of $V$ that converge to the zero vector $h_n \to 0$. Equivalently, the first order expansion holds:
$$f(x+h) = f(x) + Ah + o(h)$$
I didn't understand how the limit definition with norm got turned into the first order expansion definition. Could someone explain to me how the equivalence can be shown in more detail?
In the first order expansion $o(h)$ has to be function $o : U \to W$ such that $\lim_{h \to 0}\frac{\lVert o(h) \rVert}{\lVert h \rVert} = 0$. Intuively, it has to go faster to $0$ than $h$.
If $f$ is Frechet differentiable at $x$, then we define $$o(h) = f(x+h) - f(x) - A h .$$ Of course this is the only possibility to get $f(x+h) = f(x) + Ah + o(h)$. Then by definition $\lim_{h \to 0}\frac{\lVert o(h) \rVert}{\lVert h \rVert} = 0$.
Conversely, if we have a first order expansion $f(x+h) = f(x) + Ah + o(h)$, then $$\frac{\lVert f(x+h) - f(x) - A h \rVert}{\lVert h \rVert} = \frac{\lVert o(h) \rVert}{\lVert h \rVert}$$ which gives Frechet differentiability at $x$.