Question Number 170565 by cortano1 last updated on 27/May/22
Answered by Mathspace last updated on 27/May/22
$${f}^{'} \left({x}\right)={x}^{\mathrm{3}} \frac{\mathrm{2}\left({x}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{3}\left({x}+\mathrm{2}\right)^{\mathrm{4}} }−{x}^{\mathrm{3}} \frac{\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{3}\left({x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\Rightarrow{lim}_{{x}\rightarrow+\infty} {x}^{\mathrm{3}} {f}^{'} \left({x}\right)= \\ $$$${lim}_{{x}\rightarrow+\infty} \frac{\mathrm{2}{x}^{\mathrm{5}} }{−\mathrm{3}{x}^{\mathrm{4}} }−\frac{\mathrm{2}{x}^{\mathrm{5}} }{−\mathrm{3}{x}^{\mathrm{4}} }=\mathrm{0} \\ $$$$ \\ $$