quando occore $o(f)\subsetez o(g)$?

The definition of $o(f(x))$ implies that if $f(x) < g(x)$, then $o(f(x)) \subseteq o(g(x))$: given a function $h(x) \in o(f(x))$, eventually $|h(x)| < \epsilon \cdot f(x)$ for any $\epsilon$, but then $|h(x)| < \epsilon \cdot g(x)$ as well.

This makes intuitive sense: if $h(x)$ is insignificantly small compared to $f(x)$, then it's insignificantly small compared to larger functions as well.

However, as we'll see, sometimes $o(f(x))$ and $o(g(x))$ are the same even if $f(x) < g(x)$ for all $x$.

Different ways to write $o(f(x))$

By our definition, $o(x^2)$ is the set of functions $f(x)$ that eventually become much smaller than $x^2$: for any $\epsilon > 0$, we have $|f(x)| < \epsilon \cdot x^2$ whenever $x$ is large enough.

What about $o(10x^2)$? The definition is the same, except that we want $|f(x)| < \epsilon \cdot 10x^2 = 10\epsilon \cdot x^2$ to hold, instead. This makes the threshold $\epsilon$ weaker: using $\epsilon = 0.01$ relative to $10x^2$, for instance, is the same as using $\epsilon = 0.1$ relative to $x^2$.

But the bound $|f(x)| < \epsilon \cdot x^2$ has to hold for any $\epsilon$ we pick! If $f(x) \in o(10x^2)$, then we still know that eventually we'll have $|f(x)| < 0.01 x^2$; we just have to pick $\epsilon = 0.001$ instead of $\epsilon=0.01$. This means that $o(10x^2)$ and $o(x^2)$ are just different ways of describing the same set.

More is true. Writing $o(x^2 + 100x)$ would also describe the same set: since $100x \in o(x^2)$, eventually $x^2 + 100x$ becomes less than $x^2 + x^2$, and $o(x^2) = o(2x^2)$.

Comparing $o(f(x))$ and $o(g(x))$

In general, to have $o(f(x)) \subseteq o(g(x))$, we don't need to have $f(x) < g(x)$. All that we need is that for some (possibly large) constant $C$, we eventually have $f(x) < C\cdot g(x)$.

Intuitively, since we can make our thresholds arbitrarily small, we don't care about constant factors. And since we only care about large values of $x$, we don't care about how $f(x)$ and $g(x)$ compare for small $x$.