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Question Number 161409 by LEKOUMA last updated on 17/Dec/21
Calculate lim_(x→0) ((1−cos (1−cos x))/x^4 ) lim_(x→0) ((1−cos xcos 2xcos 3x)/x^2 )
$${Calculate} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{{x}^{\mathrm{4}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}\mathrm{cos}\:\mathrm{2}{x}\mathrm{cos}\:\mathrm{3}{x}}{{x}^{\mathrm{2}} } \\ $$
Commented by cortano last updated on 17/Dec/21
(2) lim_(x→0) ((1−cos x cos 2x cos 3x)/x^2 ) = lim_(x→0) ((1−cos x+cos x−cos x cos 2x cos 3x)/x^2 ) = lim_(x→0) ((1−cos x)/x^2 ) + lim_(x→0) ((cos x (1−cos 2x cos 3x))/x^2 ) = (1/2) + lim_(x→0) ((1−cos 2x+cos 2x−cos 2x cos 3x)/x^2 ) = (1/2)+lim_(x→0) ((1−cos 2x)/x^2 )+lim_(x→0) ((cos 2x(1−cos 3x))/x^2 ) = (1/2)+2+lim_(x→0) ((1−cos 3x)/x^2 ) = (5/2)+(9/2) = ((14)/2) = 7
$$\:\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{2}{x}\:\mathrm{cos}\:\mathrm{3}{x}}{{x}^{\mathrm{2}} } \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}+\mathrm{cos}\:{x}−\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{2}{x}\:\mathrm{cos}\:\mathrm{3}{x}}{{x}^{\mathrm{2}} } \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}^{\mathrm{2}} }\:+\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\:\mathrm{cos}\:\mathrm{3}{x}\right)}{{x}^{\mathrm{2}} } \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:\mathrm{2}{x}\:\mathrm{cos}\:\mathrm{3}{x}}{{x}^{\mathrm{2}} } \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}{{x}^{\mathrm{2}} }+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{2}{x}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{3}{x}\right)}{{x}^{\mathrm{2}} } \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{3}{x}}{{x}^{\mathrm{2}} } \\ $$$$\:=\:\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}\:=\:\frac{\mathrm{14}}{\mathrm{2}}\:=\:\mathrm{7}\: \\ $$
Answered by qaz last updated on 17/Dec/21
lim_(x→0) ((1−cos xcos 2xcos 3x)/x^2 ) =−lim_(x→0) ((ln(cos xcos 2xcos 3x))/x^2 ) =−lim_(x→0) ((lncos x)/x^2 )−lim_(x→0) ((lncos 2x)/x^2 )−lim_(x→0) ((lncos 3x)/x^2 ) =lim_(x→0) ((1−cos x)/x^2 )+lim_(x→0) ((1−cos 2x)/x^2 )+lim_(x→0) ((1−cos 3x)/x^2 ) =(1/2)+(1/2)∙2^2 +(1/2)∙3^2 =7 −−−−−−−−−−−− lim_(x→0) ((1−cos (1−cos x))/x^4 ) =lim_(x→0) (((1/2)(1−cos x)^2 )/x^4 ) =(1/2)lim_(x→0) ((((1/2)x^2 )^2 )/x^4 ) =(1/8)
$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{xcos}\:\mathrm{2xcos}\:\mathrm{3x}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=−\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{cos}\:\mathrm{xcos}\:\mathrm{2xcos}\:\mathrm{3x}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=−\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{lncos}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} }−\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{lncos}\:\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} }−\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{lncos}\:\mathrm{3x}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} }+\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} }+\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{3x}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{2}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{3}^{\mathrm{2}} \\ $$$$=\mathrm{7} \\ $$$$−−−−−−−−−−−− \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} } \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Commented by LEKOUMA last updated on 17/Dec/21
Good! Many thank
$${Good}!\:\:{Many}\:{thank} \\ $$
Answered by cortano last updated on 17/Dec/21
(1) lim_(x→0) ((1−cos (1−cos x))/x^4 ) = lim_(x→0) ((2sin^2 (((1−cos x)/2)))/x^4 ) = lim_(x→0) ((2sin^2 (((2sin^2 ((x/2)))/2)))/x^4 ) = 2×((4×(1/(16)))/4) = 2×(1/(16))=(1/8)
$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{{x}^{\mathrm{4}} } \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{2}}\right)}{{x}^{\mathrm{4}} } \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}}\right)}{{x}^{\mathrm{4}} } \\ $$$$\:\:=\:\mathrm{2}×\frac{\mathrm{4}×\frac{\mathrm{1}}{\mathrm{16}}}{\mathrm{4}}\:=\:\mathrm{2}×\frac{\mathrm{1}}{\mathrm{16}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$

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