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
$${u}\rightarrow\mathrm{0}\Rightarrow\mathrm{1}−{cosu}\:\sim\:\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}−{cos}\left(\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} }\:\: \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \:−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} } \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }−\mathrm{1}\right)}{{x}^{\mathrm{2}} } \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} −{x}^{\mathrm{4}} }{\mathrm{2}{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }= \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{2}−{x}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }=\:\mathrm{1}\:\checkmark \\ $$$$ \\ $$
Commented by cortano12 last updated on 22/Jun/23
$$\mathrm{1}\:=\:\mathrm{wrong} \\ $$
Answered by cortano12 last updated on 22/Jun/23
$$\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{23}}{\mathrm{24}} \\ $$
Commented by MM42 last updated on 22/Jun/23
$${why}? \\ $$$${please}\:{provide}\:{your}\:{solution} \\ $$