Why is the product of little ohs the little oh of the product? If we have an expression such as $o(a) \cdot o(b)$ as $a, b \to 0$ can we simplify it to $o(ab)$ as $ab \to 0$? If so how do we prove it and does it apply to general functions $o(f(a))o(g(b))=o(f(a)g(b))$ as $a, b \to 0$?
 A: By definition little-$o$ is set of functions
$$o(g)=\{f: \exists \varepsilon(x), \lim\limits_{x \to x_0}\varepsilon(x)=0,  \exists \delta >0, \forall x \in  U_\delta(x_0), f(x)= \varepsilon(x) g(x) \}$$
More exact notation for $o(g)$ often is written as $o(g(x)),x \to x_0$. From this definition we have well known property
$$o(f) \cdot o(g)=o(f \cdot g),x \to x_0$$
Are you asking about proof for this property?
As to your question: outgoing from definition $o(a), a \to a_0$ is  considering as little-$o$ for identity function: $o(g(a)), a \to a_0$, where $g(a)=a$. Same for $o(b)$.
We can consider $o(a)\cdot o(b)$ when both $a,b$ are functions with respect to one variable and then it will be $o(a(x))⋅o(b(x)),x \to x_0$. Properties $a \to 0, b \to 0$ then will be properties $a(x) \to 0, b(x) \to 0, x \to x_0$.
A: I'm not sure how qualified I am to answer your question, but from what I can tell, yes, you can simplify the expression.
I'm not perfect at proofs, but here's how I would start. I would say that if your functions are continuous, the limits as "a" and "b" approach 0 will be equal. The function will also be defined as "ab" approaches 0 given that for any a < 1, b < 1 -> ab < a & b and "ab" approaches 0. So the limit as "ab" approaches 0 will also equal the limits as "a" and "b" approach zero, and you can combine their functions into the one you mentioned, o(ab). Definitely correct me if any of that is incorrect or incomplete.
To generalize, I would say that if the limit as "a" approaches 0 of f(a) equals the limit as "b" approaches 0 of f(b), the two functions can be combined once again using the same concept I mentioned in the previous paragraph. From my understanding, you could even further simplify the last expression you gave into o(f(ab)) as a,b approach 0.
I hope this helps, although I realize it's probably not strongest proof. (Also, I don't really know how to type out functions and zero's the way you did, so I apologize for that.)
