I want to prove the following right-hand limit (one sided limit) using $\epsilon-\delta$ definition;

$\lim_{u\to 0^+} {u^{s_0} f(-ln u)} = 0$

where $f$ is a function from $R \to R$ and $s_0$ is a positive real number.

Can anybody assist me in it.

I searched out its proof or any hint, in this forum, but could not find it.

  • $\begingroup$ It may be an easy question, but I have not any clue about its proof. Can anybody run me with some proof steps... $\endgroup$ – A.A Mar 31 '16 at 7:15
  • $\begingroup$ What is $f$ ? .. $\endgroup$ – Zanzi Mar 31 '16 at 7:34
  • $\begingroup$ $f$ is any function from $R \to R$. Let me edit question please. $\endgroup$ – A.A Mar 31 '16 at 7:45
  • $\begingroup$ The existence of the limit depends on $f$ and $s_0$... With these information, nothing can be said. $\endgroup$ – Crostul Mar 31 '16 at 8:08
  • $\begingroup$ I have added the information about $s_0$ i the question. $\endgroup$ – A.A Mar 31 '16 at 8:24

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