Showing that if a function is $O(x^2)$, then it's also $o(x)$ How to prove the relationship above? The big $O$ and small $o$ are used in the context of $x\to 0$.
I've been told the proof goes like that:
If $|f(x)|<cx^2 $ (that's the property $f(x)=O(x^2)$, then of course also $\frac{|f(x)|}{|x|}<c|x|$ is arbitrarily small when $x\to 0$ (meaning $f(x)=o(x)$.
Could anyone explain why? How to prove it rigorously? What I see above is the fact that both sides of inequality were divided by a number, and that operation doesn't change the direction of inequality. But I'm not sure if it's rigorous enough.
Would $O(x^2)$ imply $o(x)$ if I'd chosen a different limit than $x\to 0$, for instance $x\to a$ for some positive $a$? I guess not, because $c |x|$ doesn't get arbitrarily small in this inequality $\frac{|f(x)|}{|x|}<c|x|$ (if it did, that would mean $f(x)$ is $o(x)$.
 A: Remember that $f \in O(x^2)$ for $x\to 0$ means that in some neighbourhood of $0$ for some constant $C$ it holds: $|f(x)| < C x^2$. While $f\in o(x)$ means that $f(x)/x \to 0$ as $x \to 0$. So, if $f\in O(x^2)$ you have
$$
\left |\frac{f(x)}{x}\right| \le C |x|
$$
in a neighbourhood of $0$. Remember now that the limit of a function in a point only depends on the values of the faction in any neighbourhood of the point. So the limit on the left exists and is equal to $0$ since the limit on the right is zero (comparison principle).
A: Taking  following definitions for $a$ point:
$$O(g) = \left\lbrace f:\exists C > 0, \exists \delta >0, \forall x( |x-a|<0) (|f(x)| \leqslant C \cdot |g(x)|) \right\rbrace$$
$$o(g) = \left\lbrace f:\exists \epsilon(x), \lim_{x \to a}\epsilon(x)=0, \exists \delta >0, \forall x( |x-a|<0) (f(x) = \epsilon (x) \cdot g(x)) \right\rbrace$$
So, both are sets and first gives some "intellectual" boundary for its $f$ elements, while second gives some kind of representation for its $f$ elements in $a$'s neighbourhood. Main relationship is:
$$o(g) \subset O(g), x \to a.$$
Proof: let's take $\phi \in o(g)$, then $\exists \epsilon(x)$ and neighbourhood $U(a)$ of $a$ in which $f(x) = \epsilon (x) \cdot g(x)$. As $\lim_{x \to a}\epsilon(x)=0$ then we can find some neighbourhood $V(a)$ in which $|\epsilon(x)| \leqslant C$, for any $C>0$. In intersection $U(a) \cap V(a)$ there will be $|f(x)| \leqslant C \cdot |g(x)|$, which finishes $\phi \in O(g)$.
Now we can consider $a=0$ and show $$O(x^2) \subset o(x), x \to 0.$$
Proof: let's take $\phi \in O(x^2)$, then we know $|\phi(x)| \leqslant C \cdot |x|^2$ for some $C>0$ and in some neighbourhood $U(0)$. Taking representation $\phi(x) = x \cdot \frac{\phi(x)}{x} = x \cdot \epsilon (x)$ for $x \in U(0) \setminus \{ 0 \}$, we obtain $\phi \in o(x)$.
