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 $o(x)$.
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).