Question Number 143043 by 0731619 last updated on 09/Jun/21
Answered by Dwaipayan Shikari last updated on 09/Jun/21
$${y}={x}!^{{x}!} \\ $$$${log}\left({y}\right)={x}!{log}\left({x}!\right) \\ $$$${log}\left({y}\right)=\Gamma\left({x}+\mathrm{1}\right){log}\left(\Gamma\left({x}+\mathrm{1}\right)\right) \\ $$$$\frac{{y}'}{{y}}=\Gamma'\left({x}+\mathrm{1}\right){log}\left(\Gamma\left({x}+\mathrm{1}\right)\right)+\Gamma'\left({x}+\mathrm{1}\right) \\ $$$${x}^{{x}^{{x}^{..} } } ={m}\Rightarrow{x}^{{m}} ={m}\Rightarrow{mlog}\left({x}\right)={log}\left({m}\right)\Rightarrow\frac{{m}'}{{m}}={m}'{log}\left({x}\right)+\frac{{m}}{{x}} \\ $$$$\left(\frac{\mathrm{1}}{{m}}−{log}\left({x}\right)\right){m}'=\frac{{m}}{{x}}\Rightarrow{m}'=\frac{{m}^{\mathrm{2}} }{\mathrm{1}−{mlog}\left({x}\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}^{{x}^{{x}^{{x}^{} } } } −{x}!}{{x}!^{{x}!} −\mathrm{1}}\overset{{Lhopital}} {=}\frac{\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{1}−\mathrm{1}{log}\left(\mathrm{1}\right)}−\Gamma'\left(\mathrm{1}+\mathrm{1}\right)}{\Gamma'\left(\mathrm{2}\right){log}\left(\Gamma\left(\mathrm{2}\right)\right)+\Gamma'\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}−\left(\mathrm{1}−\gamma\right)}{\mathrm{1}−\gamma}=\frac{\gamma}{\mathrm{1}−\gamma}\:\:\:\:\gamma={EulerMascheroni}\:{constant}\:=\mathrm{0}.\mathrm{57721} \\ $$