Simple small o exercises: $1+o(1)=\exp[o(1)]$ and $o[f(x)g(x)]=o[f(x)]O[g(x)]$

Question

(I'm self-learning these things as I am new to them.)

Prove:

(a) $1+o(1)=\exp[o(1)]$

(b) $o[f(x)g(x)]=o[f(x)]O[g(x)]$

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First, I understand that each of these statements entails a 2-way inclusion so there needs to be 2 directions for each. Is this correct? Normally, for a statement $f(x)=o(\text{something})$, I know it doesn't make sense to write $o(\text{something})=f(x)$, but in (a) and (b) above, $o$ and $O$ show up on both sides so I think these are statements about sets.

For (a), if $g(x)=o(1)$, then $$ \log[1+g(x)]=g(x)-\frac{[g(x)]^2}{2}+\frac{[g(x)]^3}{3}-\cdots=o(1)-o(1)+o(1)-\cdots=o(1). $$ Is the last "equality" above correct and why? I feel uneasy about adding an infinite number of "small" quantities and assert that the sum is also "small." I can do the reverse direction but there is the same concern.

For (b), could you please check if the basic idea below is correct? One direction: $$ h(x)=o[f(x)g(x)]\implies h(x)=\frac{h(x)}{g(x)}g(x)=o(f(x))O[g(x)] $$ where the last "equality" is because $g(x)=O[g(x)]$ trivially. The reverse direction: $$ h(x)=o(f(x)), k(x)=O[g(x)]\implies\frac{h(x)k(x)}{f(x)g(x)}=\frac{h(x)}{f(x)}\frac{k(x)}{g(x)}\to 0 $$ because one fraction goes to $0$ and the other is eventually bounded. Thus, $h(x)k(x)=o(f(x)g(x))$.

Answer 1

Firstly, it is important to note that there isn't consistent usage for big-oh or little-oh notation. But your interpretation as showing that your statements are about sets is a common and reasonable interpretation.

1. You are correct to worry about adding up an infinite number of small pieces. For example, consider $$ 1 + 1/(2x) + 1/(3x) + 1/(4x) + \cdots \approx 1 + o(1) + o(1) + o(1) + \cdots$$ but in fact this doesn't converge at all. Your proposed solution for the first part of your post doesn't work for this reason.

   With your particular application, since $g(x) = o(1)$, there is some $X$ such that for all $x > X$, we have that $\lvert g(x) \rvert < 1/2$. Restricting to $x > X$, you have that $$ 1 + g(x) - g(x)^2 / 2 \leq \log(1 + g(x)) \leq 1 + g(x),$$ from which it now follows that $\log(1 + g(x)) = 1 + o(1)$.

2. The basic idea of the second part of your post is correct.