Question Number 170809 by thean last updated on 31/May/22
Answered by som(math1967) last updated on 31/May/22
$$\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{e}^{−\mathrm{1}} \left({e}^{\mathrm{2}{x}} −\mathrm{1}\right){sin}\mathrm{3}{x}}{{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{6}}{{e}}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left({e}^{\mathrm{2}{x}} −\mathrm{1}\right)}{\mathrm{2}{x}}×\frac{{sin}\mathrm{3}{x}}{\mathrm{3}{x}} \\ $$$$=\frac{\mathrm{6}}{{e}}×\mathrm{1}=\frac{\mathrm{6}}{{e}} \\ $$
Answered by Mathspace last updated on 31/May/22
$${f}\left({x}\right)=\frac{{e}^{\mathrm{2}{x}−\mathrm{1}} −{e}^{−\mathrm{1}} }{{x}^{\mathrm{2}} }{sin}\left(\mathrm{3}{x}\right) \\ $$$${sin}\left(\mathrm{3}{x}\right)\sim\mathrm{3}{x}\:{and} \\ $$$${e}^{\mathrm{2}{x}−\mathrm{1}} −{e}^{−\mathrm{1}} ={e}^{−\mathrm{1}} \left({e}^{\mathrm{2}{x}} −\mathrm{1}\right) \\ $$$${e}^{{u}} \sim\mathrm{1}+{u}\:\Rightarrow{e}^{\mathrm{2}{x}} −\mathrm{1}\sim\mathrm{2}{x}\:\Rightarrow \\ $$$${f}\left({x}\right)\sim\frac{\mathrm{2}{xe}^{−\mathrm{1}} }{{x}^{\mathrm{2}} }×\mathrm{3}{x}\:=\frac{\mathrm{6}}{{e}} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\frac{\mathrm{6}}{{e}} \\ $$