Question Number 32517 by abdo imad last updated on 26/Mar/18
$${calculatelim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\frac{{x}^{{sinx}} \:\:−\left({sinx}\right)^{{x}} }{{x}}\:. \\ $$
Commented by abdo imad last updated on 27/Mar/18
$${we}\:{have}\:{x}^{{sinx}} \:=\:{e}^{{sinx}\:{ln}\left({x}\right)} \:{but}\:{sinx}\:\sim\:{x}\:−\frac{{x}^{\mathrm{3}} }{\mathrm{6}} \\ $$$${e}^{{sinxln}\left({x}\right)} \:\sim\:{e}^{\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right){lnx}} \:\sim\:\mathrm{1}+\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right){lnx} \\ $$$$\left({sinx}\right)^{{x}} \:={e}^{{xln}\left({sinx}\right)} \:\sim\:{e}^{{xln}\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)} \:={e}^{{xlnx}\:+{xln}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)} \\ $$$$\sim\:{e}^{{xlnx}} \:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:\sim\:{e}^{{xlnx}} \:\left(\mathrm{1}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)\sim\:\Rightarrow \\ $$$$\frac{{x}^{{sinx}} \:−\left({sinx}\right)^{{x}} }{{x}}\:\:\sim\:\:\frac{\mathrm{1}+\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right){lnx}\:−{e}^{{xlnx}} \:+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:{e}^{{xlnx}} }{{x}} \\ $$$$\sim\:\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\:{e}_{{x}\rightarrow\mathrm{0}^{+} } ^{{xln}\left({x}\right)} \:\rightarrow\mathrm{0}\:\:. \\ $$