Question Number 66696 by mathmax by abdo last updated on 18/Aug/19
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\frac{\pi}{\mathrm{4}}}{{xsin}\left({x}^{\mathrm{2}} \right)} \\ $$
Commented by kaivan.ahmadi last updated on 18/Aug/19
$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{3}{x}^{\mathrm{2}} /\left(\mathrm{1}+\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\mathrm{2}} \right)}{\mathrm{3}{x}^{\mathrm{2}} }={lim}_{{x}\rightarrow\mathrm{0}} \left(\mathrm{1}/\mathrm{1}+\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\mathrm{2}} \right)=\mathrm{1}/\mathrm{2} \\ $$
Commented by ~ À ® @ 237 ~ last updated on 18/Aug/19
$${let}\:{named}\:{it}\:{L} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\:−\frac{\pi}{\mathrm{4}}}{{x}^{\mathrm{3}} }\:.\:\frac{{x}^{\mathrm{2}} }{{sin}\left({x}^{\mathrm{2}} \right)}\: \\ $$$${Now}\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} }{{sin}\left({x}^{\mathrm{2}} \right)}\:=\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\frac{{sinu}}{{u}}}\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\left(\:{u}={x}^{\mathrm{2}} \right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\frac{\pi}{\mathrm{4}}}{{x}^{\mathrm{3}} }\:=\:\underset{{v}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{arctan}\left(\mathrm{1}+{v}\right)−\frac{\pi}{\mathrm{4}}}{{v}}\:=\underset{{v}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{1}+{v}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\left({v}={x}^{\mathrm{3}} \right) \\ $$$${So}\:\:\:\:{L}=\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 19/Aug/19
$${let}\:{f}\left({u}\right)\:={arctan}\left(\mathrm{1}+{u}\right)\:\:{we}\:{have}\: \\ $$$${f}\left({u}\right)={f}\left(\mathrm{0}\right)+{u}\:{f}^{'} \left(\mathrm{0}\right)+{o}\left({u}^{\mathrm{2}} \right)\:\:{but}\:{f}\left(\mathrm{0}\right)\:=\frac{\pi}{\mathrm{4}} \\ $$$${f}^{'} \left({u}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }\:\Rightarrow{f}^{'} \left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{f}\left({u}\right)\:=\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}{u}\:+{o}\left({u}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\:=\frac{\pi}{\mathrm{4}}\:+\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{6}} \right)\:\Rightarrow{arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\frac{\pi}{\mathrm{4}}\sim\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\:\:\left({x}\rightarrow\mathrm{0}\right) \\ $$$${slso}\:{xsin}\left({x}^{\mathrm{2}} \right)\:\sim\:{x}^{\mathrm{3}} \:\Rightarrow\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\frac{\pi}{\mathrm{4}}}{{xsin}\left({x}^{\mathrm{2}} \right)}\:={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}^{\mathrm{3}} }{\mathrm{2}{x}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$
Answered by Cmr 237 last updated on 18/Aug/19
Answered by Cmr 237 last updated on 18/Aug/19
$${with}\:{the}\:{limited}\:{develpment}\:{we}\: \\ $$$${have}: \\ $$$${arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)=\frac{\pi}{\mathrm{4}}+\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{3}} \right) \\ $$$${sin}\left({x}^{\mathrm{2}} \right)={x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right) \\ $$$${so} \\ $$$${lim}\underset{{x}\rightarrow\mathrm{0}} {\:}\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\frac{\pi}{\mathrm{4}}}{{xsin}\left({x}^{\mathrm{2}} \right)}={li}\underset{} {{m}}\underset{{x}\rightarrow\mathrm{0}} {\:}\frac{\frac{\pi}{\mathrm{4}}+\frac{{x}^{\mathrm{3}} }{\mathrm{2}}−\frac{\pi}{\mathrm{4}}}{{x}×{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}. \\ $$