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Prove that exist function $\varphi :\left( {0,\varepsilon } \right) \to \mathbb{R}$ such that $$\mathop {\lim }\limits_{x \to {0^ + }} \varphi \left( x \right) = 0,\mathop {\lim }\limits_{x \to {0^ + }} \varphi \left( x \right)\ln x = - \infty .$$ I think $\varphi \left( x \right) = \frac{1}{{\ln \left( {\ln \left( {\Gamma \left( x \right)} \right)} \right)}}$ is fine. But i can't check that. If we replace $ln$ by $f\left( x \right)$ such that $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = - \infty$, can we find $\varphi$?

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Let $f$ be such that $\lim_{x\to0+} f(x)=-\infty$ Let $\phi (x)=\frac 1 {\sqrt |f(x)|}$. Then $\phi (x) \to 0$ and $\phi (x) f(x) \to -\infty$ (because $|\phi (x) f(x)|=\sqrt {|f(x)|}$ and $\phi (x) f(x)$ has the same sign as $f(x))$. This answers the second part and the first part is a special case when $f(x)=\ln (x)$.

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    $\begingroup$ What is $f$ in this answer? $\endgroup$ – Ethan Bolker May 27 '18 at 13:44
  • $\begingroup$ @EthanBolker look at the last part of the question where $f$ is given. The first part is only a special case so my construction answers both parts. $\endgroup$ – Kavi Rama Murthy May 28 '18 at 5:37
  • $\begingroup$ Could you please add some phrase to the answer to explain a bit more. I agree it is an answer, but it is hard to see at first glance how it relates to the question. $\endgroup$ – quid May 28 '18 at 11:38
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    $\begingroup$ @quid Have you seen the edited answer? If you need more details I will be happy to help. $\endgroup$ – Kavi Rama Murthy May 29 '18 at 4:43
  • $\begingroup$ Thank you for the expansion. $\endgroup$ – quid May 29 '18 at 15:08
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I guess you want something that shrinks to zero, but does so slower than $\ln$ grows. So what about $$\varphi(x) = |\ln(x)|^{-1/2}$$ I think that should work.

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