Question Number 64011 by Prithwish sen last updated on 12/Jul/19
$$\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\frac{\mathrm{x}^{\mathrm{x}} −\mathrm{1}}{\mathrm{xlnx}} \\ $$
Commented by kaivan.ahmadi last updated on 12/Jul/19
$${y}={x}^{{x}} \Rightarrow{lny}={xlnx}\Rightarrow\frac{{y}'}{{y}}={lnx}+{x}×\frac{\mathrm{1}}{{x}}=\mathrm{1}+{lnx}\Rightarrow \\ $$$${y}'={y}\left(\mathrm{1}+{lnx}\right)={x}^{{x}} \left(\mathrm{1}+{lnx}\right) \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \frac{{x}^{{x}} −\mathrm{1}}{{xlnx}}={lim}_{{x}\rightarrow\mathrm{0}^{+} } \frac{{x}^{{x}} −\mathrm{1}}{{lnx}^{{x}} }\:\:\overset{{hop}} {=} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\frac{{x}^{{x}} \left(\mathrm{1}+{lnx}\right)}{\frac{{x}^{{x}} \left(\mathrm{1}+{lnx}\right)}{{x}^{{x}} }}={lim}_{{x}\rightarrow\mathrm{0}^{+} } {x}^{{x}} = \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } {e}^{{xlnx}} ={e}^{\mathrm{0}} =\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 12/Jul/19
$${let}\:{A}\left({x}\right)=\frac{{x}^{{x}} −\mathrm{1}}{{xlnx}}\:\Rightarrow{A}\left({x}\right)=\frac{{e}^{{xln}\left({x}\right)} −\mathrm{1}}{{xlnx}}\:\:\:{we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{xln}\left({x}\right)=\mathrm{0} \\ $$$${and}\:{e}^{{u}} \:\sim\:\mathrm{1}+{u}\:\:\:\left({u}\in{V}\left(\mathrm{0}\right)\right)\:\Rightarrow{e}^{{xln}\left({x}\right)} =\mathrm{1}+{xln}\left({x}\right)\:\Rightarrow \\ $$$${e}^{{xln}\left({x}\right)} −\mathrm{1}\:\sim{xln}\left({x}\right)\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow\frac{{e}^{{xln}\left({x}\right)} −\mathrm{1}}{{xlnx}}\:\sim\mathrm{1}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{A}\left({x}\right)\:=\mathrm{1}\:. \\ $$
Commented by Prithwish sen last updated on 12/Jul/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sirs}. \\ $$
Answered by Rio Michael last updated on 12/Jul/19
$$\left({lnx}+\mathrm{1}\right){x}^{{x}} \\ $$$${let}\:{y}=\:{x}^{{x}} \:\:\:{since}\:\frac{{d}\left(−\mathrm{1}\right)}{{dx}}=\mathrm{0} \\ $$$${lny}\:=\:{lnx}^{{x}} \\ $$$$\:\frac{\mathrm{1}}{{y}}=\:{x}\left(\frac{\mathrm{1}}{{x}}\right)+\:{lnx} \\ $$$$\:\:\frac{\mathrm{1}}{{y}}=\:{lnx}\:+\:\mathrm{1} \\ $$$$\:\:\:\:=\:\left({lnx}+\mathrm{1}\right){x}^{{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left({lnx}+\mathrm{1}\right){x}^{{x}} }{{lnx}+\mathrm{1}} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{x}^{{x}} \\ $$$$=\:\mathrm{0}\:\:\: \\ $$$${please}\:{check}\: \\ $$
Commented by Prithwish sen last updated on 12/Jul/19
$$\mathrm{thamk}\:\mathrm{you}\:\mathrm{sir} \\ $$