Question Number 71360 by 20190927 last updated on 14/Oct/19
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{2x}^{\mathrm{4}} }\mathrm{cos}\:\left(\sqrt{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{5}} \mathrm{ln}\:\left(\mathrm{1}−\mathrm{2x}^{\mathrm{3}} \right)} \\ $$
Commented by mathmax by abdo last updated on 14/Oct/19
$${for}\:{x}\in\:{V}\left(\mathrm{0}\right)\:\:\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{4}} }\:\sim\mathrm{1}+{x}^{\mathrm{4}} \:\:\:{and}\:{cos}\left(\sqrt{\mathrm{2}}{x}^{\mathrm{2}} \right)\sim\mathrm{1}−\frac{\mathrm{2}{x}^{\mathrm{4}} }{\mathrm{2}}\:=\mathrm{1}−{x}^{\mathrm{4}} \:\Rightarrow \\ $$$$\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{4}} }\:{cos}\left(\sqrt{\mathrm{2}}{x}^{\mathrm{2}} \right)\sim\mathrm{1}−\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\left(\mathrm{1}−{x}^{\mathrm{4}} \right)=\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{8}} \right)\:={x}^{\mathrm{8}} \\ $$$${ln}\left(\mathrm{1}−\mathrm{2}{x}^{\mathrm{3}} \right)\:\sim−\mathrm{2}{x}^{\mathrm{3}} \:\Rightarrow{x}^{\mathrm{5}} {ln}\left(\mathrm{1}−\mathrm{2}{x}^{\mathrm{3}} \right)\sim−\mathrm{2}{x}^{\mathrm{8}} \:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{4}} }{cos}\left(\sqrt{\mathrm{2}}{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{5}} {ln}\left(\mathrm{1}−\mathrm{2}{x}^{\mathrm{3}} \right)}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by 20190927 last updated on 17/Oct/19
$$\mathrm{thanks} \\ $$