$h(x) = o(f(x)g(x)) \implies h(x) = f(x) \cdot o(g(x))$? Let's define $o(g(x))$ as usually:
$$
\forall x \ne a.g(x) \ne 0 \\
f(x) = o(g(x)) \space \text{when} \space x \to a \implies \lim_{x \to a} \frac{f(x)}{g(x)}=0
$$
Is it true that: $h(x) = o(f(x)g(x)) \implies h(x) = f(x) \cdot o(g(x))$?
The way I would approach proving that is as follows:
$$
h(x) = o(f(x)g(x)) \text{ when $x \to a$ } \implies \lim_{x \to a} \frac{h(x)}{f(x)g(x)} = 0 \implies \lim_{x \to a} \frac{\frac{h(x)}{f(x)}}{g(x)} = 0  \\
\implies \frac{h(x)}{f(x)} = o(g(x)) \implies h(x) = f(x) \cdot o(g(x))
$$
I know the implication in the other way holds (question), but I'm not sure if it is bidirectional (if the two right hand-sides are equivalent). For example, if $h(x) = o(x) \implies h(x) = x \cdot o(1)$, the right hand-side of the right side of the implication is defined at $x = 0$, while the right hand side of the left side of the implication is not. That would confuse me if that equivalence of the two sides holds because we should completely ignore the point $0$ when analysing that?
Thanks!
 A: First of all, a disclaimer: your notation is incorrect.
$o(f)$ is not a function, it is a family of functions. In particular, $F\in o(f)$ is true if and only if $$\lim_{x\to a} \frac{F(x)}{f(x)}$$ is true.
With that, you may notice that the notation $h(x)=o(f(x)g(x))$ is a little strange, because you have what you want is a function on the left, but a set on the right. So it would be better to phrase your question as

If $h\in o(f\cdot g)$, does that mean that $h=f\cdot G$ for some $G\in o(g)$?

and this is the question I will address from no on.

The answer is yes.
You can first write $h(x)=f(x)\cdot G(x)$ where
$$G(x)=\begin{cases}
\frac{h(x)}{f(x)};&f(x)\neq 0\\
0;&f(x)=0\end{cases}$$
now all you need to do is prove that $G\in o(g)$. This is true because
$$\frac{G(x)}{g(x)} = \begin{cases}\frac{h(x)}{f(x)g(x)};&f(x)\neq 0\\ 0;&f(x)=0\end{cases}.$$
which means that it should be easy to see, for example from the sandwich theorem or just by definition of limit, that $$\lim_{x\to a} \frac{G(x)}{g(x)} = 0$$
