"Of the order of" notation I have a function where all terms have the same coefficient $x^3$ in it:
For example
$f(x) = ax^3 - bx^3$
Can I say that in big $O$ notation:


*

*$f(x) = O(x^3)$ 

*$f(x) = O(x^3)$ as $x \rightarrow 0$

*I cannot say that.
Also, Can I say that in little $o$ notation:
$f(x) = o(x^2)$ as $x \rightarrow 0$?
 A: To start with: whenever you want to use asymptotics, always specify where the asymptotics are taken. I.e., it is when $x\to a$, for some $a\in\mathbb{R}\cup\{\pm\infty\}$: it may be clear in context, but if not you need to state it (typically, $0$ or $\infty$). This rules out 1., just because it's ambiguous (although, as we will see, if you did specify where, it'd be correct in your case).
Now, for your specific example: for any $x\in\mathbb{R}$, by the triangle inequality
$$
\lvert f(x)\rvert = \lvert a-b\rvert \lvert x^3\rvert \leq 
(\lvert a\rvert+\lvert b\rvert) \lvert x^3\rvert \tag{$\dagger$}
$$
so, by definition of the $O(\cdot)$ notation (take constant $C>0$ equal to $\lvert a\rvert+\lvert b\rvert$, or even $\lvert a\rvert+\lvert b\rvert +1$ to discard the case $a=b=0$) we do have
$$
f(x) = O(x^3) 
$$
when $x\to 0$. (For that matter, also when $x\to \infty$, or even  $x\to a$ for any $a$). So 2. is correct (regardless of the values of $a,b$).
Note: if $a\neq b$, you even have the stronger statement $f(x) = \Theta(x^3)$ when $x\to 0$ (or, again, $x\to a$ for any $a$).
Second question: can you write $f(x) = o(x^2)$ at $0$? Again, let us check with the definition:

$f(x) = o(x^2)$ when $x\to 0$ if there exists a non-negative function $\varepsilon$ such that $\lvert f(x) \rvert = \varepsilon(x)\lvert g(x)\rvert$ and $\varepsilon(x)\xrightarrow[x\to 0]{} 0$.

Well, let's use $(\dagger)$ again, with $\varepsilon(x)\stackrel{\rm def}{=} (\lvert a\rvert+\lvert b\rvert) \lvert x\rvert$. This satisfies both conditions, so that indeed 
$$
f(x) = o(x^2)
$$
when $x\to 0$. (Now, you can check that this will not be true when $x\to \infty$, unless $a=b$. In this case, $f=0$, and this is trivially true.)
