big and small O notation help

It isn't. But if $f(x)$ is O$(x^3)$ as $x\to0$, then $|f(x)|\lt C|x^3|$ for some positive constant $C$, so $|f(x)/x^2|\lt C|x|$, so $f(x)/x^2\to0$ as $x\to0$, so $f(x)$ is o$(x^2)$.

Recall that the rigorous meaning of the Bachmann-Landau notation [$f=O(g)$ at $x_0$] is that $f\in O_{x_0}(g)$ where $O_{x_0}(g)$ is the set of functions $h$ such that there exists a finite $C$ and a neighborhood $V$ of $x_0$ such that $g$ and $h$ are defined at least on $V\setminus\{x_0\}$ and $|h(x)|\leqslant C\cdot|g(x)|$ for every $x$ in $V\setminus\{x_0\}$.

Similarly, $o_{x_0}(g)$ is the set of functions $h$ such that there exists a function $k$ and a neighborhood $V$ of $x_0$ such that $g$, $h$ and $k$ are defined at least on $V\setminus\{x_0\}$, $h=kg$ on $V\setminus\{x_0\}$ and $\lim\limits_{x\to x_0}\,k(x)=0$.

Now, let $x_0=0$ and consider the functions $u$ and $v$ defined by $u(x)=x^3$ and $v(x)=x^2$ for every $x$. Then it is true that $O_{x_0}(u)\subseteq o_{x_0}(v)$ in the sense that every function $w$ in $O_{x_0}(u)$ belongs to $o_{x_0}(v)$. But $o_{x_0}(v)\not\subseteq O_{x_0}(u)$ since for example the function $w$ defined by $w(x)=|x|^{5/2}$ is in $o_{x_0}(v)$ but not in $O_{x_0}(u)$.

To sum up, $f=O(g)$ at $x_0$ and $f=o(g)$ at $x_0$, often written as $f(x)=O(g(x))$ at $x_0$ and $f(x)=o(g(x))$ at $x_0$ respectively, are extremely convenient shorthands for $f\in O_{x_0}(g)$ and $f\in o_{x_0}(g)$ respectively, but nothing more.