Question Number 93287 by john santu last updated on 12/May/20
Commented by mathmax by abdo last updated on 12/May/20
$${let}\:{f}\left({x}\right)\:=\frac{{arctan}\left({x}^{\mathrm{2}} \left(\mathrm{1}−{cosx}\right)\right)}{\left(\mathrm{1}−\sqrt{{cosx}}\right){ln}\left(\frac{{sinx}}{{x}}\right)}{we}\:{have}\:\mathrm{1}−{cosx}\:\sim\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow{x}^{\mathrm{2}} \left(\mathrm{1}−{cosx}\right)\sim\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\:\Rightarrow \\ $$$${arctan}\left({x}^{\mathrm{2}} \left(\mathrm{1}−{cosx}\right)\right)\:\sim{arctan}\left(\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\right)\sim\:\frac{{x}^{\mathrm{4}} }{\mathrm{2}} \\ $$$${sinx}\:\sim{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow\frac{{sinx}}{{x}}\sim\:\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}\:\Rightarrow{ln}\left(\frac{{sinx}}{{x}}\right)\sim{ln}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}\right)\sim−\frac{{x}^{\mathrm{2}} }{\mathrm{6}} \\ $$$${cosx}\:\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\sqrt{{cosx}}\sim\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow\mathrm{1}−\sqrt{{cosx}}\sim\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:\sim\frac{\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}×\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}\right)}\:=−\frac{\mathrm{4}×\mathrm{6}}{\mathrm{2}}\:=−\mathrm{12}\:\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:{f}\left({x}\right)\:=−\mathrm{12} \\ $$$$ \\ $$