$o(x)$ and $O(x^2)$

Question

How to prove, that $O(x^2) \subset o(x) $ when $x \to 0$? How should i use the definitions of Big O : $ \exists C>0, \exists \delta : |x|<\delta, |f(x)| \leq C|x^2| $ and little o: $\forall \epsilon>0 \> \exists \delta>0 : 0<|x|<\delta, |f(x)| \leq \epsilon|x| $ if the first definition mentions two constants and the second has $\forall \epsilon>0$?

Answer 1

If $f(x) = O(x^2)$, then there are $c>0$ and $d > 0$ such that $f(x) < cx^2$ for $|x| < d$.

Therefore, for $|x| < d$, $\frac{f(x)}{x} < c|x|$.

You want to show that, for any $\delta > 0$, there is an $\epsilon > 0$ such that $|\frac{f(x)}{x}| < \delta $ for $|x| < \epsilon $.

In $\frac{f(x)}{x} < c|x|$, to make $\frac{f(x)}{x} < \epsilon$, choose $c|x| < \epsilon$, or $|x| < \epsilon/c$.

Therefore, if $|x| < \min(d, \epsilon/c)$. $\frac{f(x)}{x} < \epsilon$.

Answer 2

You can use the following fact: if $g(x)\neq 0$ on an interval $I$ centered around $0$, and $\frac{f(x)}{g(x)} \xrightarrow[x\to 0]{} C > 0$, then $f(x) = O(g(x))$. Similarly, if $\frac{f(x)}{g(x)} \xrightarrow[x\to 0]{} 0$, then $f(x) = o(g(x))$

Answer 3

Hint:

First, write out what it means for $f(x)=O(x^2)$; that is, there is a $C$ so that $|f(x)|\le Cx^2$ for $x$ near $0$.

Next, $f(x)=o(x)$ means that $\lim\limits_{x\to0}\frac{f(x)}{x}=0$.