$f(x) = o(g(x)) \implies f(x) = o(o(g(x)))$ and $o(g(x)) = o(o(g(x)))$ Let's define $o(g(x))$ as usually:
$$
\forall x \ne a.g(x) \ne 0 \\
f(x) = o(g(x)) \space \text{when} \space x \to a \implies \lim_{x \to a} \frac{f(x)}{g(x)}=0
$$
So for any continuous function $g$:
$o(g(x)) = o(o(g(x))) \implies \lim_{x \to a} \frac{o(g(x))}{o(g(x))} = 0$, but it should be $1$. Because of the contradiction, we can conclude that the initial statement is false.
As far as I can see, $f(x) = o(g(x)) \implies f(x) = o(o(g(x)))$ holds:
\begin{equation}
f(x) = o(g(x)) \text{ when } x \to a \implies \lim_{x \to a} \frac{f(x)}{g(x)} = 0 \\
\implies \lim_{x \to a} \frac{f(x)}{g(x)} \cdot 0 = 0 \\
\implies \lim_{x \to a} \frac{f(x)}{g(x)} \cdot \lim_{x \to a} \frac{o(f(x))}{f(x)} = 0  \\
\implies \lim_{x \to a} \frac{f(x)}{g(x)} \cdot \frac{o(f(x))}{f(x)} = 0 \\
\implies \lim_{x \to a} \frac{o(f(x))}{g(x)} = 0
\end{equation}
\begin{equation}
\tag{By hypothesis $f(x) = o(g(x))$}
\implies \lim_{x \to a} \frac{o(o(g(x)))}{g(x)} = 0
\end{equation}
\begin{equation}
\tag{1}
\implies o(o(g(x))) = o(g(x))
\end{equation}
Is it true that, even though by hypothesis $f(x) = o(g(x))$, we cannot say that $o(o(g(x))) = f(x)$ in the equation $1$, because $f(x)$ is a concrete function from the set $o(g(x))$, and it could be that $f(x)$ does not satisfy the equation $1$, while a different element of $o(g(x))$ does (e.g. $=$ there has a different meaning than usual equals sign, so it could be even considered as a notation abuse)? Is that a valid explanation why $[f(x) = o(g(x)) \implies f(x) = o(o(g(x)))] \not \implies o(g(x)) = o(o(g(x)))$?
I'm still getting the handle on proofs with limits, and I'm looking for the second pair of eyes to make sure I'm not making a mistake...
Thanks!
 A: Your question is based on a misunderstanding of the $o$-notation. When we say for instance $f(x)=o(g(x))$, this is a kind of abuse of notation, as you suspect. It looks like we are saying that $f(x)$ is equal to something, but that is not what it means at all $-$ it is a shorthand notation for the meaning that you give at the start of your post, and is not an equation in the usual sense of the word.
$o(g(x))$ can be used as an argument to some function on the right-hand side of an equation; for instance $f(x)=e^{o(g(x))}$. But it can't occur on the left-hand side of an equation, and it can't be applied to itself. So both $o(g(x))=f(x)$ and $f(x)=o(o(g(x)))$ are meaningless $-$ their meaning hasn't been defined.
Your chain of reasoning breaks down, as far as I can see, when you write
$$\implies \lim_{x \to a} \frac{f(x)}{g(x)} \cdot \lim_{x \to a} \frac{o(f(x))}{f(x)} = 0$$
Here, $o(f(x))$ occurs on the left-hand side of an equation, so it is meaningless.
See this question and its comments for a related case.
