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I would like to show that $$ (\ln\ln(x) )(\ln x )+\ln\ln\ln x=o_{+\infty} (\ln x )^{2}+\ln\ln\ln\ln x $$

by using these methods :

  • limit $$\lim_{x\to +\infty }\dfrac{ (\ln\ln(x) )(\ln x )+\ln\ln\ln x }{ (\ln x )^{2}+\ln\ln\ln\ln x }$$

  • properties of symbol o $$ \ln x = o(x) $$ enter image description here

enter image description here

$$\dfrac{ (\ln\ln(x) )(\ln x )+\ln\ln\ln x }{ (\ln x )^{2}+\ln\ln\ln\ln x }=\dfrac{\ln\ln x +\dfrac{ \ln\ln\ln x }{ \ln x } }{ \ln x +\dfrac{ \ln\ln\ln\ln x }{ \ln x } } $$ in this last expression how can i use $ \lim_{ x\to +\infty }\dfrac{ \ln x }{x} =0$

Educ
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    What did the drowning number theorist say? "log log log log log log log" – Jacob Claassen May 22 '17 at 05:49
  • yes this helps to see why the limit goes to 0 thanks, but could we use second method which means properties of small o as first photo shows – Educ May 22 '17 at 09:58

1 Answers1

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Hint. By letting $x=\ln(t)$, $$\lim_{ x\to +\infty }\dfrac{ \ln x }{x} =0\implies \lim_{ t\to +\infty }\dfrac{ \ln(\ln t) }{\ln(t)}=0.$$ Hence your limit is $$\lim_{ x\to +\infty }\dfrac{\frac{\ln\ln x}{\ln x} +\dfrac{ \ln\ln\ln x }{ (\ln x)^2 } }{ 1 +\dfrac{ \ln\ln\ln\ln x }{ (\ln x)^2 } }=0$$

Robert Z
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