Proof of Property of little o I am learning some asymptotic analysis and i'm trying to prove:
If $\lim_{x\to c} f(x)< \infty$ then $o(fg)=o(g)$ (as $x \to c$). So far i have proved that $o(fg)\subset o(g)$, but i am stuck on the other set inclusion. Is it possible that this is not true? I saw the same claim for Big O instead so I thought it might be true for little o since $o(f)\subset O(f)$ but this could be bad intuition...
so my questions are: is the property true? If so how can i show the 2nd set inclusion? Any hints are greatly appreciated. If wanted i can add my work that shows the first set inclusion. Thank you.
(Note: all functions are defined on some domain with $c$ being a limit point of the common domain and i am using the definition that $o(g)$ is the class of all functions $f(x)$ such that for all $\epsilon >0$ there exists $\delta >0$ so that
 $$|f(x)| < \epsilon |g(x)| $$ for any $x$ in the domain of $f(x)$ and $g$ and $0<|x-c|<\delta$)
Edit:
I believe it is trivial if i allow myself to use the limit definition of $o(g)$ for nonzero $g$ by the following argument:
if $\lim_{x\to c} f(x)=L$ then if $h\in o(g)$ by definition of $o(g)$ we have $\lim_{x\to c} h(x)/g(x)=0$ and so
$$\lim_{x\to c} \frac{h(x)}{f(x)g(x)}=1/L \cdot 0=0$$
(Provided $f(x)$ is never zero and L is not zero)

It's not true. What happens is $f(x) = 0$ for all $x$? It's not true for Big-O either, for exactly the same reason. As your edit shows, you have successfully proven it when the limit is not $0$.
