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)

@Glinka To edit "epsilon" into "varepsilon" in a 3 years old question is petty at best. Please spend your energy on worthier causes. — Did, Jun 26 '16 at 17:06
@Did, because it looks better this way. What motivates you to reverse it manually is far more interesting question — Glinka, Jun 26 '16 at 17:11
@Did, I say this. In my opinion it looks better. The question of something "looking better" in general is subjective. At my understanding \epsilon is shorter to type and it saves time, but \varepsilon is better percepted. So whenever I see the first way of writing the letter, I correct to the second one. You're saying I shouldn't? — Glinka, Jun 26 '16 at 17:22
@Glinka Rhetorical question if ever there was one (or cannot you read my first comment?). — Did, Jun 26 '16 at 17:26
@Did, don't understand your hostility... Or why you went through trouble of correcting it back. You don't explain your action at all. But whatever, the matter is paltry and I'll trust you know better. I'll keep in mind not to make such edits again. — Glinka, Jun 26 '16 at 17:45

score 2 · Accepted Answer · edited Jun 26 '16 at 17:08

2

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 $f(\epsilon)=\epsilon$

edited Jun 26 '16 at 17:08

Did

279,727

answered May 28 '13 at 21:36

Isn't $f=o(1)$ ? – Belgi May 28 '13 at 21:39
@Belgi Yes, it is. But $f(\epsilon)/\epsilon=1$ is not. – Pedro May 28 '13 at 21:41
My mistake, for a moment I forgot I need to divide $f$ by $\epsilon$. Thanks for the answer! – Belgi May 28 '13 at 21:43
@Belgi In my counterexample $f=o(1)$ but $\frac{f(\epsilon)}{\epsilon}\neq o(1)$ – May 28 '13 at 21:43
@Belgi You're welcome. – May 28 '13 at 21:50

score 1 · Answer 2 · answered May 28 '13 at 21:36

1

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? :-)

answered May 28 '13 at 21:36

Peter Košinár

10,810

Acually, I planned on taking $f=\epsilon\ln(\epsilon)\cdot c(\epsilon$. So this gives us Sami's answer, which turns to be a valid counterexample. But it seems you think your answer is not useful to the original question..why ? – Belgi May 28 '13 at 21:45
I originally misread the lower-case $o(1)$ as upper-case $O(1)$ in which case you wouldn't get a contradiction (since $42=O(1)$). – Peter Košinár May 28 '13 at 21:59

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

2 Answers2