Does $f(\epsilon)=o(\epsilon\ln(\epsilon))$ imply $\frac{f(\epsilon)}{\epsilon}=o(1)$?

Question

I have the following homework question:

Does $f(\epsilon)=o(\epsilon\ln(\epsilon))$ imply $\frac{f(\epsilon)}{\epsilon}=o(1)$ ?

It doesn't seem correct to me, using the definition I could only get $$\frac{f(\epsilon)}{\epsilon}=o(\ln(\epsilon))$$ and it doesn't seem that if $g(\epsilon)=o(\ln(\epsilon))$ then $g(\epsilon)=o(1)$.

To contradict this I wish to find a function $c(\epsilon)$ s.t $$\lim_{\epsilon\to0}c(\epsilon)=0$$ but $$\lim_{\epsilon\to0}|c(\epsilon)\ln(\epsilon)|>0$$

I am having difficulty finding such a function, I tried a few and checked them using WA, they all satisfied $$\lim_{\epsilon\to0}c(\epsilon)=0$$ but they all also satisfied that $$\lim_{\epsilon\to0}|c(\epsilon)\ln(\epsilon)|=0$$

Can someone please help me find such $c(\epsilon)$ ? (or surprise me and show that this statement is in fact true)

Answer 1

The notation $$f(x)=_a o(g(x))$$ means $$\lim_{x\to a}\frac{f(x)}{g(x)}=0\iff \frac{f(x)}{g(x)}=_a o(1) $$ and to contradict your implication just take $f(\epsilon)=\epsilon$

Answer 2

Be careful what you wish for -- setting $c(x)=\frac{42}{\ln x}$ satisfies your conditions, but doesn't help much with the original question... or does it? :-)