Question Number 136596 by 0731619177 last updated on 23/Mar/21
Answered by MJS_new last updated on 24/Mar/21
![answer is (γ/(1−γ)) lim_(x→1) ((x^x^(x...) −xΓ(x))/(xΓ(x)^(xΓ(x)) −1)) = =lim_(x→1) (((d/dx)[x^x^(x...) −xΓ(x)])/((d/dx)[xΓ(x)^(xΓ(x)) −1])) y=x^x^(x...) ⇔ ln y =yln x ⇒ y′=(y^2 /(x(1−yln x))) Γ′(1)=−γ this should be easy to show](https://www.tinkutara.com/question/Q136663.png)
$$\mathrm{answer}\:\mathrm{is}\: \\ $$$$\frac{\gamma}{\mathrm{1}−\gamma} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{{x}^{{x}…} } −{x}\Gamma\left({x}\right)}{{x}\Gamma\left({x}\right)^{{x}\Gamma\left({x}\right)} −\mathrm{1}}\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\left[{x}^{{x}^{{x}…} } −{x}\Gamma\left({x}\right)\right]}{\frac{{d}}{{dx}}\left[{x}\Gamma\left({x}\right)^{{x}\Gamma\left({x}\right)} −\mathrm{1}\right]} \\ $$$${y}={x}^{{x}^{{x}…} } \:\Leftrightarrow\:\mathrm{ln}\:{y}\:={y}\mathrm{ln}\:{x}\:\Rightarrow\:{y}'=\frac{{y}^{\mathrm{2}} }{{x}\left(\mathrm{1}−{y}\mathrm{ln}\:{x}\right)} \\ $$$$\Gamma'\left(\mathrm{1}\right)=−\gamma \\ $$$$\mathrm{this}\:\mathrm{should}\:\mathrm{be}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{show} \\ $$