Could anyone explain to me how did we get this result? This is simple example but I'm struggling with little o notation in general.

Question

$\left(x-\displaystyle\frac{x^3}{6}+\displaystyle\frac{x^5}{120} +o(x^5)\right)\left(1+\displaystyle\frac{x^2}{2}-\displaystyle\frac{x^4}{24}+ \displaystyle\frac{x^4}{8}+o(x^4)\right) = x+\displaystyle\frac{x^3}{3}+\displaystyle\frac{2x^5}{15}+o(x^5)$

Answer 1

Expand the product in the usual way, and then (I guess the small $o$ is to be meant near $0$) use the following rules (you can prove them as an easy exercise by using the definition of small $o$):

1. $x^k=o(x^n)$ for $k>n$
2. $x^ko(x^n)=o(x^{k+n})$
3. $o(x^k)o(x^n)=o(x^{k+n})$
4. $o(x^k)=o(x^n)$ for $k\ge n$

V.I.: this notation is NOT symmetric: the meaning is that -take the first one for example- $$ \frac{x^k}{x^n}\stackrel{x\to0}{\longrightarrow}0. $$

EDIT The notation $o(x^n)$ denotes the class of all real continuous functions defined, say, on a open interval near $0$, such that they are going to $0$ faster than $x\mapsto x^n$, that is: $$ \lim_{x\to0}\frac{f(x)}{x^n}=0\;. $$

To understand this, I would suggest to make some example: $x^3$ is $o(x^2)$ but is NOT $o(x^3)$ and so on.