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Question Number 113524 by Khalmohmmad last updated on 13/Sep/20

Answered by Dwaipayan Shikari last updated on 13/Sep/20

$${tan}^{-1}(x^2-x^4)=y$$$$tany=x^2(1-x^2)x\to \mathrm{\infty }$$$$tany\to -\mathrm{\infty }$$$$y=-\frac{\pi }{2}$$

Answered by mathmax by abdo last updated on 13/Sep/20

$$\mathrm{the}\mathrm{function}\mathrm{x}\to \mathrm{arctanx}\mathrm{is}\mathrm{increazing}\mathrm{on}\mathrm{so}$$$${\lim }_{\mathrm{x}\to \mathrm{\infty }}{\mathrm{x}}^2-{\mathrm{x}}^4={\lim }_{\mathrm{x}\to \mathrm{\infty }}-{\mathrm{x}}^4=-\mathrm{\infty }\Rightarrow {\lim }_{\mathrm{x}\to \mathrm{\infty }}\arctan ({\mathrm{x}}^2-{\mathrm{x}}^4)$$$$={\lim }_{\mathrm{x}\to \mathrm{\infty }}\arctan (-{\mathrm{x}}^4)=-\frac{\pi }{2}$$

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