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Question-160816




Question Number 160816 by 0731619 last updated on 07/Dec/21
Commented by mr W last updated on 07/Dec/21
f(t)=t^t^t^(...)   =((W(−ln t))/(−ln t)) is defined only  for t>0.  with t=ln x, so (ln x)^((ln x)^(...) )  is only   defined for ln x>0 or x>1.  therefore lim_(x→0)  (ln x)^((ln x)^(...) )  doesn′t  exist.
$${f}\left({t}\right)={t}^{{t}^{{t}^{…} } } =\frac{{W}\left(−\mathrm{ln}\:{t}\right)}{−\mathrm{ln}\:{t}}\:{is}\:{defined}\:{only} \\ $$$${for}\:{t}>\mathrm{0}. \\ $$$${with}\:{t}=\mathrm{ln}\:{x},\:{so}\:\left({ln}\:{x}\right)^{\left({ln}\:{x}\right)^{…} } \:{is}\:{only}\: \\ $$$${defined}\:{for}\:\mathrm{ln}\:{x}>\mathrm{0}\:{or}\:{x}>\mathrm{1}. \\ $$$${therefore}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{ln}\:{x}\right)^{\left(\mathrm{ln}\:{x}\right)^{…} } \:{doesn}'{t} \\ $$$${exist}. \\ $$

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