<s>[INST] Algebra of o-symbols In theorem $7.8$, Tom Apostol in his Calculus Vol. $1$ gave and proved the following basic rules for algebra of o-symbols

*

*$o(g(x)) \pm o(g(x)) = o(g(x)) $

*$o(cg(x)) = o(g(x)) = o(g(x)) $, if $c \ne 0$

*$f(x) \cdot o(g(x)) = o(f(x)g(x)) $

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

*$\frac{1}{1 + g(x)} = 1 - g(x) + o(g(x))$
Now when proving that $\tan x = x + \frac{1}{3}x^3 + o(x^3)$ as $x \to 0$, he used the following equality, only mentioning that $g(x) = -\frac{1}{2}x^2 + o(x^3)$ and that he is applying the 5th case of the theorem above:
$$
Equation 1: \frac{1}{1 - \frac{1}{2}x^2 + o(x^3)} = 1 + \frac{1}{2}x^2 + o(x^2)
$$
By just applying that 5th case of the theorem, we get
$$
Equation 2: \frac{1}{1 - \frac{1}{2}x^2 + o(x^3)} = 1 + \frac{1}{2}x^2 - o(x^3) + o(-\frac{1}{2}x^2 + o(x^3))
$$
Now, I can prove that the right handside of Eq1 is equal to the right hand side of Eq2 (e.g. using this method), by proving that:

*

*$o(-\frac{1}{2}x^2 + o(x^3)) = o(x^2)$

*$o(x^2) + o(x^3) = o(x^2)$
But I wonder if there is an obvious reason why the above two equalities directly follow from the Theorem $7.8$?

Edit, to add a suggested proof for  $o(x^2) + o(x^3) = o(x^2)$, which I hope was not even necessary, because it could be inferred trivially from the theorem $7.8$, potentially by the application of step $4$ $n-m$ times.
Let $m < n \land f(x) = o(x^m) \land g(x) = o(x^n)$.
Then we have:
$f(x) + g(x) = o(x^m) + o(x^n) = o(x^m) \cdot (1 + x^{n - m})$, because of T7.8:3.
\begin{equation}
\implies  \frac{f(x) + g(x)}{1 + x^{n - m}} = o(x^m)  \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \cdot (1 + 0) \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \cdot (\lim_{x \to 0} 1 + \lim_{x \to 0} x^{n-m})
\end{equation}
The last equality holds because $\lim_{x \to 0} x^{n-m} = 0$. Then further we have:
\begin{equation}
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \cdot (\lim_{x \to 0} \frac{x^m + x^n}{x^m}) \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \cdot \frac{x^m + x^n}{x^m} \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m} \\
\implies f(x) + g(x) = o(x^m)
\end{equation}
The last equality follows by definition of little-o. By just substituting $f$ and $g$, we get $o(x^m) + o(x^n) = o(x^m)$.
 [/INST] The best way to deal with the $"o"$ notation is to write it as
$$F(x)=o(g(x)) \; (x\to 0)$$
means that
$$F(x)=g(x)\epsilon(x)$$
with
$$\lim_{x\to 0}\epsilon(x)=0$$
So, your last equality gives
$$o(x^2)+o(x^3)=$$
$$x^2\epsilon_1(x)+x^3\epsilon_2(x)=$$
$$x^2(\epsilon_1(x)+x\epsilon_2(x))=$$
$$x^2\epsilon(x)=o(x^2)$$
</s>