Property of little o notation $f(x) o (g(x)) = o(f(x)g(x))$ This question is regarding property of little o notation given in Apostol Calculus. The property is given on page 288 and stated as:

Theorem 7.8 (c) As $x\to a$ we have $f(x)\cdot o (g(x)) = o(f(x)g(x))$.

Here say $h(x) = o(g(x))$ then we have $f(x) \lim_{x\to a} \frac{h(x)}{g(x)} = 0$, and on the right, we have $j(x) = o(f(x)g(x)) $ then $\lim_{x\to a} \frac{j(x)}{f(x)g(x)} = 0$ .. I am confused to approach the proof. 
 A: By definition
$$f(x) o (g(x)) = f(x)\cdot \omega(x)g(x)$$ 
with
$$\omega(x)\to 0 \quad x\to a$$
therefore
$$f(x) o (g(x)) = \omega(x)\cdot f(x)g(x)=o(f(x)g(x))$$
A: $$o(g(x))=g(x)\epsilon(x)$$ with $$\lim_{x\to a}\epsilon(x)=0$$
then
$$f(x)o(g(x))=\Bigl(f(x)g(x)\Bigr)\epsilon(x)=o(f(x)g(x))$$
A: This answer is based upon Apostol's settings. First of all we should be aware that the equality  sign $=$ here is used to indicate a subset relation $\subseteq$ between two sets 
\begin{align*}
f(x)o(g(x))=o(f(x)g(x)\qquad\qquad \text{as }x\to a
\end{align*}
meaning that whenever a function  $h=h(x)$ is an element of  the left-hand  side $f(x)o(g(x))$ it is  also  an element  of  the right-hand  side $o(f(x)g(x))$ as $x\to a$.

We obtain
  \begin{align*}
\color{blue}{h(x)}&\color{blue}{=f(x)o(g(x))\qquad\text{as }x\to a}\\
&\quad:\Longleftrightarrow\ \exists h_1: h(x)=f(x)h_1(x)\quad\text{ and }\quad\lim_{x\to a}\frac{h_1(x)}{g(x)}=0\\
&\quad\ \ \Longrightarrow\ \exists h_1: h(x)=f(x)h_1(x)\quad\text{ and }\quad\lim_{x\to a}\frac{f(x)h_1(x)}{f(x)g(x)}=0\\
&\quad\ \ \Longrightarrow\ \ \color{blue}{h(x)=o(f(x)g(x))\qquad \text{as }x\to a}
\end{align*}
  and the claim follows.

