Question Number 142309 by qaz last updated on 29/May/21
$$\underset{\mathrm{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{x}^{\left(\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{x}} } −\left(\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{x}^{\mathrm{sin}\:\mathrm{x}} } }{\mathrm{x}^{\mathrm{3}} }=? \\ $$
Answered by mnjuly1970 last updated on 29/May/21
$$\:\:\:{lim}_{{x}\rightarrow\mathrm{0}} \:\left({sin}\left({x}\right)\right)^{{x}} ={lim}_{{x}\rightarrow\mathrm{0}} \left({x}\right)^{{sin}\left({x}\right)} =\mathrm{1} \\ $$$$\:\:{A}:={lim}_{{x}\rightarrow\mathrm{0}} \left({sin}\left({x}\right)\right)^{{x}} \\ $$$$\:\:\:{log}\left({A}\right)={lim}_{{x}\rightarrow\mathrm{0}} {xlog}\left({sin}\left({x}\right)\right)\overset{{hop}} {=}\mathrm{0} \\ $$$$\:\:\:\:{A}={e}^{\mathrm{0}} =\mathrm{1} \\ $$$$\:\:\:\:\:{similarly}\:\:{lim}_{{x}\rightarrow\mathrm{0}} \left({x}\right)^{{sin}\left({x}\right)} =\mathrm{1} \\ $$$$\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{{x}−{sin}\left({x}\right)}{{x}^{\mathrm{3}} }\underset{{expansion}} {\overset{{maclaurin}} {=}}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{{x}−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\right)}{{x}^{\mathrm{3}} } \\ $$$$\:\:\:\boldsymbol{\phi}\::=\:\frac{\mathrm{1}}{\mathrm{6}}\:....\checkmark \\ $$