I am trying to understand the following question: Show that $f(x)=O(|x-x_0|^2)$ as $x\to x_0$ implies $f(x)=o(|x-x_0|)$ as $x\to x_0$, but give an example to show that the converse is not true.
Definitions: We say $f(x)=O(g(x))$ as $x \to x_0$ if there exists $\frac{1}{n}$ and a positive constant c such that $|x-x_0|<\frac{1}{n}$ implies $|f(x)|≤c|g(x)|$. We say $f(x)=o(g(x))$ as $x \to x_0$ if for every $\frac{1}{m}$ there exists $\frac{1}{n}$ such that $|x-x_0|<\frac{1}{n}$ implies $|f(x)|≤|g(x)|/m$.
My attempt: We have that for $f(x)=O(|x-x_0|^2)$ as $x \to x_0$, there exists $\frac{1}{n}$ and a positive constant c such that $|x-x_0|<\frac{1}{n}$ implies $|f(x)|≤c||x-x_0|^2|<c\frac{1}{n^2}$. I want to show that $f(x)=o(|x-x_0|)$ as $x \to x_0$ if for every $\frac{1}{m}$ there exists $\frac{1}{n_1}$ such that $|x-x_0|<\frac{1}{n_1}$ implies $|f(x)|≤||x-x_0||/m<\frac{1}{mn_1}$. From the assumption I only know f(x) is bounded as $n^2|f(x)|<c$ for some n and c. What bugs me is m. Pick $n=n_1$. Suppose $\frac{1}{m}≤\frac{1}{n}$, so $\frac{1}{mn}≤\frac{1}{n^2}$. Hence, for $m≥n$, $|f(x)|≤||x-x_0||/m<\frac{1}{mn}≤\frac{1}{n^2}$, so $|f(x)|≤\frac{1}{n^2}$. Suppose, $\frac{1}{m}>\frac{1}{n}$, so $\frac{1}{mn}>\frac{1}{n^2}$. We have from the assumption that $|f(x)|≤c||x-x_0|^2|<c\frac{1}{n^2}<c\frac{1}{mn}$. So, we have shown for every $\frac{1}{m}$ there exists $\frac{1}{n}$, such that $|x-x_0|<\frac{1}{n}$ implies $|f(x)|≤||x-x_0||/m<\frac{1}{mn}$.
Is this allowed? Is this the way these questions should be tackled?
For the converse question, I am thinking that I want to show that there exists $\frac{1}{n}$ and a positive constant c such that $|x-x_0|<\frac{1}{n}$ implies $|f(x)|>c||x-x_0|^2|$. I am not sure if this is right, but I want to let $m=n^2$, so that then $|f(x)|≤||x-x_0||/m<|x-x_0|^3$. But I get stuck here, and I do not know how I can show that the implication is false in the converse direction.
Thanks in advance!
$f(x)=O(|x-x_0|^2)$ as $x\rightarrow x_0$ means that there is a constant $C>0$ and a neighborhood $U$ of $x_0$ such that $$|f(x)|\leq C|x-x_0|^2,\qquad x\in U$$ Hence $$\frac{|f(x)|}{|x-x_0|}\leq C|x-x_0|,\qquad x\in U\setminus\{x_0\}$$ The conclusion follows now directly by letting $x\rightarrow x_0$.