Menu Close

lim-x-0-1-1-2x-4-cos-2-x-2-x-5-ln-1-2x-3-




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

Leave a Reply

Your email address will not be published. Required fields are marked *