Question Number 113524 by Khalmohmmad last updated on 13/Sep/20
Answered by Dwaipayan Shikari last updated on 13/Sep/20
$${tan}^{−\mathrm{1}} \left({x}^{\mathrm{2}} −{x}^{\mathrm{4}} \right)={y} \\ $$$${tany}={x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:\:\:\:{x}\rightarrow\infty \\ $$$${tany}\rightarrow−\infty \\ $$$${y}=−\frac{\pi}{\mathrm{2}} \\ $$
Answered by mathmax by abdo last updated on 13/Sep/20
$$\mathrm{the}\:\mathrm{function}\:\mathrm{x}\rightarrow\mathrm{arctanx}\:\mathrm{is}\:\mathrm{increazing}\:\mathrm{on}\:\mathrm{so}\: \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \mathrm{x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{4}} \:=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} −\mathrm{x}^{\mathrm{4}} \:=−\infty\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \mathrm{arctan}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{4}} \right) \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \:\mathrm{arctan}\left(−\mathrm{x}^{\mathrm{4}} \right)\:=−\frac{\pi}{\mathrm{2}} \\ $$