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lim-x-0-1-cos-2x-1-n-x-2-1-3-n-




Question Number 159395 by cortano last updated on 16/Nov/21
 lim_(x→0)  ((1−((cos 2x))^(1/n) )/x^2 ) = (1/3)   n=?
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt[{{n}}]{\mathrm{cos}\:\mathrm{2}{x}}}{{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:{n}=? \\ $$
Answered by qaz last updated on 16/Nov/21
lim_(x→0) ((1−((cos 2x))^(1/n) )/x^2 )=−(1/n)lim_(x→0) ((lncos 2x)/x^2 )=−(1/n)lim_(x→0) ((cos 2x−1)/x^2 )=(2/n)  ⇒n=6
$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt[{\mathrm{n}}]{\mathrm{cos}\:\mathrm{2x}}}{\mathrm{x}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{lncos}\:\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:\mathrm{2x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{n}=\mathrm{6} \\ $$
Answered by tounghoungko last updated on 16/Nov/21
 lim_(x→0)  ((1−((cos 2x))^(1/n) )/x^2 ) = (1/3)   lim_(x→0)  ((1−((1−2sin^2 x))^(1/n) )/x^2 ) = (1/3)   lim_(x→0)  ((1−(1−((2sin^2 x)/n)))/x^2 ) = (1/3)   lim_(x→0)  (((2sin^2 x)/n)/x^2 ) = (1/3) ⇒(2/n)= (1/3)   ⇒ n = 6 .
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt[{{n}}]{\mathrm{cos}\:\mathrm{2}{x}}}{{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt[{{n}}]{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}}}{{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{2sin}\:^{\mathrm{2}} {x}}{{n}}\right)}{{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2sin}\:^{\mathrm{2}} {x}}{{n}}}{{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\frac{\mathrm{2}}{{n}}=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\Rightarrow\:{n}\:=\:\mathrm{6}\:. \\ $$

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