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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)

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  • $\begingroup$ @Glinka To edit "epsilon" into "varepsilon" in a 3 years old question is petty at best. Please spend your energy on worthier causes. $\endgroup$
    – Did
    Commented Jun 26, 2016 at 17:06
  • $\begingroup$ @Censi LI Why did you approve this edit? $\endgroup$
    – Did
    Commented Jun 26, 2016 at 17:07
  • $\begingroup$ @William Why did you approve this edit? $\endgroup$
    – Did
    Commented Jun 26, 2016 at 17:07
  • $\begingroup$ @Did, because it looks better this way. What motivates you to reverse it manually is far more interesting question $\endgroup$
    – Glinka
    Commented Jun 26, 2016 at 17:11
  • $\begingroup$ @Glinka Who says so ("looks better this way")? The OP? $\endgroup$
    – Did
    Commented Jun 26, 2016 at 17:12

2 Answers 2

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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 $f(\epsilon)=\epsilon$

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  • $\begingroup$ Isn't $f=o(1)$ ? $\endgroup$
    – Belgi
    Commented May 28, 2013 at 21:39
  • $\begingroup$ @Belgi Yes, it is. But $f(\epsilon)/\epsilon=1$ is not. $\endgroup$
    – Pedro
    Commented May 28, 2013 at 21:41
  • $\begingroup$ My mistake, for a moment I forgot I need to divide $f$ by $\epsilon$. Thanks for the answer! $\endgroup$
    – Belgi
    Commented May 28, 2013 at 21:43
  • $\begingroup$ @Belgi In my counterexample $f=o(1)$ but $\frac{f(\epsilon)}{\epsilon}\neq o(1)$ $\endgroup$
    – user63181
    Commented May 28, 2013 at 21:43
  • $\begingroup$ @Belgi You're welcome. $\endgroup$
    – user63181
    Commented May 28, 2013 at 21:50
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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? :-)

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  • $\begingroup$ 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 ? $\endgroup$
    – Belgi
    Commented May 28, 2013 at 21:45
  • $\begingroup$ I originally misread the lower-case $o(1)$ as upper-case $O(1)$ in which case you wouldn't get a contradiction (since $42=O(1)$). $\endgroup$ Commented May 28, 2013 at 21:59

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