I understand that $\lim\limits_{x \to +\infty} \frac{f(x)}{g(x)} = 0$ implies that, for sufficiently large values of $x$, $f(x)<g(x)$, as a direct consequence of the definition of limit to $\infty$.
But how can the idea that $g$ grows faster than $f$ be formalized? Is it somehow related to the derivatives of the functions?
If you use Landau's symbols $$o(\cdot),\ \sim,\ \asymp$$ the limit is precisely the definition.
More specifically: $$\begin{align} f = o(g) &\iff \lim_{x \to \alpha}\frac{f(x)}{g(x)} = 0\\ f \sim_\alpha g &\iff \lim_{x \to \alpha}\frac{f(x)}{g(x)} = 1 \iff f(x) = g(x) + o(g(x))\\ f \asymp_\alpha &\iff \lim_{x \to \alpha}\frac{f(x)}{g(x)} = l \in \mathbb R,\mbox{ with }l \neq 0 \end{align}$$
Note that when using Landau's $o(\cdot)$, $\alpha$ is implicit: therefore you'll have to specify it in order to avoid confusion.
Landau's little $o$ is very useful as it offers a neat notation that has nice properties: $$\begin{align} f\cdot o(g) &= o(f\cdot g)\\ k\cdot o(f) &= o(k\cdot f) = o(f),\quad\mbox{with }k \in \mathbb R \backslash \{0\}\\ o(o(f)) &= o(f)\\ o(f + o(f)) &= o(f) \end{align}$$ You can prove all of them (and there are a few more) with the definition of $o(\cdot)$. As you can imagine, it would be very tedious to write all of them in their limit form. This is why this concise notation is often preferred.