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Question Number 136596 by 0731619177 last updated on 23/Mar/21

Answered by MJS_new last updated on 24/Mar/21

$$\mathrm{answer}\mathrm{is}$$$$\frac{\gamma }{1-\gamma }$$$$\underset{x\to 1}{\lim }\frac{x^{x^{x...}}-x\mathrm{\Gamma }\left(x\right)}{x\mathrm{\Gamma }{\left(x\right)}^{x\mathrm{\Gamma }\left(x\right)}-1}=$$$$=\underset{x\to 1}{\lim }\frac{\frac{d}{dx}[x^{x^{x...}}-x\mathrm{\Gamma }\left(x\right)]}{\frac{d}{dx}[x\mathrm{\Gamma }{\left(x\right)}^{x\mathrm{\Gamma }\left(x\right)}-1]}$$$$y=x^{x^{x...}}\iff \ln y=y\ln x\Rightarrow y^{\prime }=\frac{y^2}{x(1-y\ln x)}$$$${\mathrm{\Gamma }}^{\prime }\left(1\right)=-\gamma$$$$\mathrm{this}\mathrm{should}\mathrm{be}\mathrm{easy}\mathrm{to}\mathrm{show}$$

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