Proof of a little-o property

Question

Little-o has the following property:

$$o(o(g(x))) = o(g(x))$$

I prove it like this. Let $f_1(x) = o(g(x)), f_2(x) = o(f_1(x))$. Then the statement $o(o(g(x)))$ implies $\lim_{x \rightarrow 0} \frac{f_2(x)}{f_1(x)} = 0$, and $\lim_{x \rightarrow 0} \frac{f_1(x)}{g(x)} = 0$. We need to prove that $\lim_{x \rightarrow 0} \frac{f_2(x)}{g(x)} = 0$.

We can use $$\lim \frac{f_2(x)}{g(x)} = \lim \frac{f_2(x)}{f_1(x)}\frac{f_1(x)}{g(x)} = \lim \frac{f_2(x)}{f_1(x)}\lim \frac{f_1(x)}{g(x)} = 0\;\; (*)$$ (all limits as $x \rightarrow 0$).

Is this a valid proof? The issue here is that the division by $f_1$ in the denominator of $(*)$ might be prohibited, since it can be that $f_1(x) = 0$ as $x \rightarrow 0$. Otherwise, is there anything else wrong with this proof?

Answer 1

Recall that in Apostol's Calculus I (mentioned in comment by OP), the little o notation is defined as follows (page 286)

Whenever one writes $o(h(x))$ (as $x\to a$), it should be understood by context that $h(x)\neq 0$ for all $x\neq a$ in some interval containing $a$. So the statement $$ o(o(g(x)))=o(g(x)),\quad \textrm{as }x\to a $$ should be understood as follows:

Assume that $g(x)\neq 0$ for all $x\neq a$ in some interval containing $a$. Suppose $f(x)=o(g(x))$ such that $f(x)\neq 0$ for all $x\neq a$ in some interval containing $a$. Then $o(f(x))=o(g(x))$.

So when you write $f_2(x)=o(f_1(x))$ (as $x\to 0$), you are implicitly assuming by definition of the little o notation that $f_1(x)\neq 0$ for all $x\neq0$ in some interval containing $0$.