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Question-175122




Question Number 175122 by Michaelfaraday last updated on 20/Aug/22
Answered by Ar Brandon last updated on 20/Aug/22
L=lim_(x→0) ((1−((cos(2021x)))^(1/(2022)) )/x^2 )       =lim_(x→0) ((1−(1−((2021^2 x^2 )/2))^(1/(2022)) )/x^2 )       =lim_(x→0) (1/x^2 )(1−(1−((2021^2 x^2 )/(4044))))=((2021^2 )/(4044))
$$\mathscr{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt[{\mathrm{2022}}]{\mathrm{cos}\left(\mathrm{2021}{x}\right)}}{{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{2021}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2022}}} }{{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{2021}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{4044}}\right)\right)=\frac{\mathrm{2021}^{\mathrm{2}} }{\mathrm{4044}} \\ $$

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