Ok. in this case we are able to show that [\limsup_{z\to 0^+}\frac{g(x)}{-x\ln x}=\limsup_{n\to\infty}\frac{g(M^{-n})}{M^{-n}\ln M^{-n}}] for any $M>1$ and the same is true for the lower limit

Another problem which is connected with this limit - is the following equality true for $g$, which fullfills the above assumptions [ \limsup_{x\to 0^+}\frac{g(x)}{-x\ln x}= \sup_{M\in\mathbb{N}}\limsup_{n\to\infty}\frac{g(M^{-n})}{M^{-n}\ln M^{-n}}.] The crucial assuption seems to be concavity of $g$ (of course if the equality is true)

Pietro, thank You very much for suggestion - it has solved many problems

Let just $A=[0,1]$

Thank You very much for suggestion and for to drawing my attention to this paper