Question Number 75553 by aliesam last updated on 12/Dec/19
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{1}}{{sin}\left({x}\right)}−\frac{\mathrm{3}}{{sin}\left(\mathrm{3}{x}\right)}\right) \\ $$
Commented by mathmax by abdo last updated on 12/Dec/19
$${f}\left({x}\right)=\frac{{sin}\left(\mathrm{3}{x}\right)−\mathrm{3}{sinx}}{{x}\:{sin}\left({x}\right){sin}\left(\mathrm{3}{x}\right)}\:\:{we}\:{have}\:{sinx}=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\Rightarrow \\ $$$${sinx}\:={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+{o}\left({u}^{\mathrm{5}} \right)\:\Rightarrow{sin}\left(\mathrm{3}{x}\right)=\mathrm{3}{x}−\frac{\left(\mathrm{3}{x}\right)^{\mathrm{3}} }{\mathrm{3}!}\:+{o}\left({x}^{\mathrm{5}} \right) \\ $$$$=\mathrm{3}{x}−\frac{\mathrm{27}\:{x}^{\mathrm{3}} }{\mathrm{3}.\mathrm{2}}\:+{o}\left({x}^{\mathrm{5}} \right)\:=\mathrm{3}{x}−\frac{\mathrm{9}{x}^{\mathrm{3}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{5}} \right)\:\:\Rightarrow \\ $$$${sin}\left(\mathrm{3}{x}\right)−\mathrm{3}{sinx}\:\sim\mathrm{3}{x}−\frac{\mathrm{9}{x}^{\mathrm{3}} }{\mathrm{2}}−\mathrm{3}{x}\:+\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{5}} \right)\:=−\mathrm{4}{x}^{\mathrm{3}} \:+{o}\left({x}^{\mathrm{5}} \right) \\ $$$${and}\:{xsin}\left({x}\right){sin}\left(\mathrm{3}{x}\right)\sim\mathrm{3}{x}^{\mathrm{3}} \:\Rightarrow{f}\left({x}\right)\:\sim−\frac{\mathrm{4}}{\mathrm{3}}\:+{o}\left({x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:{f}\left({x}\right)=\frac{−\mathrm{4}}{\mathrm{3}} \\ $$
Commented by aliesam last updated on 12/Dec/19
$${thank}\:{you}\:{sir}\: \\ $$
Commented by abdomathmax last updated on 13/Dec/19
$${you}\:{are}\:{welcome}\:{sir}. \\ $$