Prove $o(x^2) = o(x^2 + o(x^3))$ when $x \to 0$ Let's define $o(g(x))$ as usually:
$$
\forall x \ne a.g(x) \ne 0 \\
f(x) = o(g(x)) \space \text{when} \space x \to a \implies \lim_{x \to a} \frac{f(x)}{g(x)}=0
$$
How to prove that $o(x^2) = o(x^2 + o(x^3)) \space \text{when} \space x \to 0$?
Could I do it like this:
$$
\text{Let} \space f(x) = o(x^2 + o(x^3)) \implies \\
0 = \lim_{x \to 0} \frac{f(x)}{x^2 + o(x^3)} = \lim_{x \to 0} \frac{f(x)}{x^2 + x^3 \frac{o(x^3)}{x^3}} = \lim_{x \to 0} \frac{f(x)}{x^2 + 0} = \lim_{x \to 0} \frac{f(x)}{x^2} \implies f(x) = o(x^2)
$$
My main question there is if the multiplication of $o(x^3)$ with $\frac{x^3}{x^3}$ is allowed?  Is there a better way to do it?
Thanks!
 A: Multiplication by $x^3/x^3$ is fine since $x\to 0$ allows us to assume $x\ne 0$ so $x^3/x^3$ is defined and equal to $1$. The problem with your proof is your substitution of the expression in the denominator with $0$. There's no theorem which allows you to do something like this and doing so is in general wrong. Here's a simple example:
$$1=\lim_{x\to 0}\frac{2x}{x+x}=\lim_{x\to 0}\frac{2x}{x+0}=\lim_{x\to 0}\frac{2x}{x} $$
which is wrong since the last limit is $2$. You can use the various limit properties but you cannot arbitrarily make these kinds of substitutions. To prove the assertion, we can write
$$\frac{f(x)}{x^2}=\frac{f(x)}{x^2+o(x^3)}\frac{x^2+o(x^3)}{x^2}=\frac{f(x)}{x^2+o(x^3)}\left(1+x\frac{o(x^3)}{x^3}\right) $$
We have $\lim_{x\to 0}x=0$ and $\lim_{x\to 0}o(x^3)/x^3=0$ so
$$\lim_{x\to 0}\left(1+x\frac{o(x^3)}{x^3}\right)=1+0=1 $$
Thus, if we assume that $f(x)=o(x^2+o(x^3))$, then
$$\lim_{x\to 0}\frac{f(x)}{x^2}=\lim_{x\to 0}\frac{f(x)}{x^2+o(x^3)}\cdot\lim_{x\to 0}\left(1+x\frac{o(x^3)}{x^3}\right)=0\cdot 1=0 $$
You can do the other direction in the same way.
