Question Number 36843 by Penguin last updated on 06/Jun/18
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({x}^{{x}^{{x}^{…..} } } \right) \\ $$
Commented by maxmathsup by imad last updated on 06/Jun/18
$${let}\:{put}\:\:{A}_{{n}} \left({x}\right)={x}^{{x}^{{x}…} } \left({n}×\right) \\ $$$${A}_{\mathrm{2}} \left({x}\right)={x}^{{x}} \:={e}^{{xln}\left({x}\right)} \\ $$$${A}_{\mathrm{3}} \left({x}\right)={x}^{{x}^{{x}} } \:=\left({A}_{\mathrm{2}} \left({x}\right)\right)^{{x}} \:={e}^{{x}^{\mathrm{2}} {ln}\left({x}\right)} \:\:\Rightarrow\:{A}_{{n}} \left({x}\right)=\:{e}^{{x}^{{n}−\mathrm{1}} {ln}\left({x}\right)} \:\:{let}\:{suppose}\:{that} \\ $$$${A}_{{n}+\mathrm{1}} \left({x}\right)=\:{x}^{{x}^{…} } \left({n}+\mathrm{1}\right)×=\left({A}_{{n}} \left({x}\right)\right)^{{x}} =\:\left({e}^{{x}^{{n}−\mathrm{1}} {ln}\left({x}\right)} \right)^{{x}} \:=\:{e}^{{x}^{{n}} {ln}\left({x}\right)} \\ $$$${we}\:{have}\:{lim}_{{x}\rightarrow+\infty\:{and}\:{n}\rightarrow+\infty} \:{A}_{{n}} \left({x}\right)\:=+\infty\:. \\ $$
Answered by MJS last updated on 06/Jun/18
$$\infty,\:\mathrm{what}\:\mathrm{else}\:\mathrm{would}\:\mathrm{you}\:\mathrm{expect}? \\ $$
Commented by maxmathsup by imad last updated on 06/Jun/18
$${that}\:{must}\:{be}\:+\infty \\ $$