Question Number 89540 by M±th+et£s last updated on 17/Apr/20
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{cos}\left({x}^{\mathrm{2}} \right)−\mathrm{1}+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}{{x}^{\mathrm{2}} \left({x}−{sin}\left({x}\right)\right)^{\mathrm{2}} } \\ $$
Commented by abdomathmax last updated on 17/Apr/20
$${let}\:{f}\left({x}\right)=\frac{{cos}\left({x}^{\mathrm{2}} \right)−\mathrm{1}+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}{{x}^{\mathrm{2}} \left({x}−{sinx}\right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:{cos}\left({u}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{u}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$=\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}!}\:+\frac{{u}^{\mathrm{4}} }{\mathrm{4}!}\:+…\Rightarrow{cos}\left({x}^{\mathrm{2}} \right)\sim\mathrm{1}−\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\:+\frac{{x}^{\mathrm{8}} }{\mathrm{4}!}\:\:\:\left({x}\:\sim\mathrm{0}\right) \\ $$$${sinx}\:\sim{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow{x}−{sinx}\:\sim\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow \\ $$$${f}\left({x}\right)\sim\frac{\mathrm{1}−\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\:+\frac{{x}^{\mathrm{8}} }{\mathrm{4}!}−\mathrm{1}+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}{{x}^{\mathrm{2}} ×\frac{{x}^{\mathrm{6}} }{\mathrm{6}^{\mathrm{2}} }}\:=\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{4}!}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:{f}\left({x}\right)=\frac{\mathrm{36}}{\mathrm{4}.\mathrm{3}.\mathrm{2}}\:=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 17/Apr/20
$${god}\:{bless}\:{you}\:{sir} \\ $$
Commented by abdomathmax last updated on 18/Apr/20
$${you}\:{are}\:{welcome}\:{sir}. \\ $$