lim-x-0-1-1-2-x-2-cos-x-1-x-2-x-4-

Question Number 193863 by horsebrand11 last updated on 21/Jun/23

$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} −\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\right)}{\mathrm{x}^{\mathrm{4}} }\:=? \\ $$

Answered by MM42 last updated on 21/Jun/23

u→0⇒1−cosu ∼ (1/2)u^2 ⇒lim_(x→0) ((1−cos((x/(1−x^2 )))−(1/2)x^2 )/x^4 ) =lim_(x→0) (((1/2)((x/(1−x^2 )))^2 −(1/2)x^2 )/x^4 ) =lim_(x→0) (((1/2)((1/((1−x^2 )^2 ))−1))/x^2 ) =lim_(x→0) ((2x^2 −x^4 )/(2x^2 (1−x^2 )^2 ))= =lim_(x→0) ((2−x^2 )/(2(1−x^2 )^2 ))= 1 ✓

$${u}\rightarrow\mathrm{0}\Rightarrow\mathrm{1}−{cosu}\:\sim\:\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}−{cos}\left(\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} }\:\: \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \:−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} } \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }−\mathrm{1}\right)}{{x}^{\mathrm{2}} } \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} −{x}^{\mathrm{4}} }{\mathrm{2}{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }= \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{2}−{x}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }=\:\mathrm{1}\:\checkmark \\ $$$$ \\ $$

Commented by cortano12 last updated on 22/Jun/23