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Question Number 143043 by 0731619 last updated on 09/Jun/21

Answered by Dwaipayan Shikari last updated on 09/Jun/21

$$y=x{!}^{x!}$$$$log\left(y\right)=x!log(x!)$$$$log\left(y\right)=\mathrm{\Gamma }(x+1)log\left(\mathrm{\Gamma }(x+1)\right)$$$$\frac{y^{\prime }}{y}={\mathrm{\Gamma }}^{\prime }(x+1)log\left(\mathrm{\Gamma }(x+1)\right)+{\mathrm{\Gamma }}^{\prime }(x+1)$$$$x^{x^{x^{..}}}=m\Rightarrow x^m=m\Rightarrow mlog\left(x\right)=log\left(m\right)\Rightarrow \frac{m^{\prime }}{m}=m^{\prime }log\left(x\right)+\frac{m}{x}$$$$(\frac{1}{m}-log\left(x\right))m^{\prime }=\frac{m}{x}\Rightarrow m^{\prime }=\frac{m^2}{1-mlog\left(x\right)}$$$$\underset{x\to 1}{\lim }\frac{x^{x^{x^{x^{}}}}-x!}{x{!}^{x!}-1}\stackrel{Lhopital}{=}\frac{\frac{1^2}{1-1log\left(1\right)}-{\mathrm{\Gamma }}^{\prime }(1+1)}{{\mathrm{\Gamma }}^{\prime }\left(2\right)log\left(\mathrm{\Gamma }\left(2\right)\right)+{\mathrm{\Gamma }}^{\prime }\left(2\right)}$$$$=\frac{1-(1-\gamma )}{1-\gamma }=\frac{\gamma }{1-\gamma }\gamma =EulerMascheroniconstant=0.57721$$

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