Question Number 78682 by Omer Alattas last updated on 19/Jan/20
Commented by mathmax by abdo last updated on 19/Jan/20
$${we}\:{have}\:\left({sinx}\right)^{{x}} ={e}^{{xln}\left({sinx}\right)} \:\:{and}\:{xln}\left({sinx}\right)\sim{xlnx}\:\:\left({x}\rightarrow\mathrm{0}\right)\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{xln}\left({sinx}\right)\:=\mathrm{0}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\left({sinx}\right)^{{x}} \:={e}^{\mathrm{0}} \:=\mathrm{1} \\ $$
Commented by john santu last updated on 20/Jan/20
$${let}\:{y}\:=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left(\mathrm{sin}\:{x}\right)^{{x}} \\ $$$${ln}\left({y}\right)=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{x}\:{ln}\left(\mathrm{sin}\:{x}\right) \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{{ln}\left(\mathrm{sin}\:{x}\right)}{{x}^{−\mathrm{1}} }\:=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{cot}\:{x}}{−{x}^{−\mathrm{2}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{−{x}^{\mathrm{2}} }{\mathrm{tan}\:{x}}\:=\:\mathrm{0} \\ $$$${finally}\:{we}\:{get}\:{e}^{{ln}\left({y}\right)} ={e}^{\mathrm{0}} \:\Rightarrow{y}\:=\:\mathrm{1} \\ $$