Question Number 160625 by Jamshidbek last updated on 03/Dec/21
$$\underset{{x}\rightarrow\mathrm{0}+\mathrm{0}} {\mathrm{lim}}\left(\mathrm{x}^{\mathrm{x}} −\mathrm{1}\right)\mathrm{lnx} \\ $$
Answered by Mathspace last updated on 03/Dec/21
$${f}\left({x}\right)=\left({x}^{{x}} −\mathrm{1}\right){lnx}\:\Rightarrow \\ $$$$\left.{f}\left({x}\right)=_{{lnx}={t}} \:\:\:\left({e}^{{t}} \right)^{{e}^{{t}} } −\mathrm{1}\right){t}=\left({e}^{{te}^{{t}} } −\mathrm{1}\right){t} \\ $$$$\left({t}\rightarrow−\infty\right) \\ $$$${g}\left({t}\right)={t}\left({e}^{{te}^{{t}} } −\mathrm{1}\right)\:={t}\:{e}^{{te}^{{t}} } −{t} \\ $$$${lim}_{{t}\rightarrow−\infty} {te}^{{te}^{{t}} } \:=\mathrm{0}\:\:{because}\:{e}^{{t}} \:{defeat}\:{t}^{{n}} \:\Rightarrow \\ $$$${lim}_{{t}\rightarrow−\infty} {g}\left({t}\right)=\mathrm{0}\:={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{f}\left({x}\right) \\ $$