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Question Number 161747 by cortano last updated on 22/Dec/21
lim_(x→0) ((cos^3 (2x)−cos (x))/(cos^2 (4x)−cos (2x))) =?
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:^{\mathrm{3}} \left(\mathrm{2}{x}\right)−\mathrm{cos}\:\left({x}\right)}{\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{4}{x}\right)−\mathrm{cos}\:\left(\mathrm{2}{x}\right)}\:=? \\ $$
Answered by Ar Brandon last updated on 22/Dec/21
L=lim_(x→0) ((cos^3 (2x)−cos(x))/(cos^2 (4x)−cos(2x)))=lim_(x→0) (((1−((4x^2 )/2))^3 −(1−(x^2 /2)))/((1−((16x^2 )/2))^2 −(1−((4x^2 )/2)))) =lim_(x→0) (((1−3((4x^2 )/2)+3((16x^4 )/4)−((64x^6 )/8))−(1−(x^2 /2)))/((1−2((16x^2 )/2)+((256x^4 )/4))−(1−((4x^2 )/2)))) =lim_(x→0) ((−((11x^2 )/2)+12x^4 −8x^6 )/(−14x^2 +64x^4 ))=lim_(x→0) ((−((11)/2)+12x^2 −8x^4 )/(−14+64x^2 ))=((11)/(28))
$$\mathscr{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}^{\mathrm{3}} \left(\mathrm{2}{x}\right)−\mathrm{cos}\left({x}\right)}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{4}{x}\right)−\mathrm{cos}\left(\mathrm{2}{x}\right)}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{3}} −\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\left(\mathrm{1}−\frac{\mathrm{16}{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} −\left(\mathrm{1}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}}\right)} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}−\mathrm{3}\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}\frac{\mathrm{16}{x}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{64}{x}^{\mathrm{6}} }{\mathrm{8}}\right)−\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\left(\mathrm{1}−\mathrm{2}\frac{\mathrm{16}{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{256}{x}^{\mathrm{4}} }{\mathrm{4}}\right)−\left(\mathrm{1}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}}\right)} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\frac{\mathrm{11}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{12}{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{6}} }{−\mathrm{14}{x}^{\mathrm{2}} +\mathrm{64}{x}^{\mathrm{4}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\frac{\mathrm{11}}{\mathrm{2}}+\mathrm{12}{x}^{\mathrm{2}} −\mathrm{8}{x}^{\mathrm{4}} }{−\mathrm{14}+\mathrm{64}{x}^{\mathrm{2}} }=\frac{\mathrm{11}}{\mathrm{28}} \\ $$
Answered by blackmamba last updated on 22/Dec/21
lim_(x→0) ((cos^3 (2x)−1+1−cos (x))/(cos^2 (4x)−1+1−cos (2x))) = lim_(x→0) ((3(cos 2x−1)+2sin^2 ((x/2)))/(2(cos (4x)−1)+2sin^2 (x))) = ((3(−2)+(1/2))/(2(−2)(4)+2)) =((−((11)/2))/(−14)) = ((11)/(28))
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:^{\mathrm{3}} \left(\mathrm{2}{x}\right)−\mathrm{1}+\mathrm{1}−\mathrm{cos}\:\left({x}\right)}{\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{4}{x}\right)−\mathrm{1}+\mathrm{1}−\mathrm{cos}\:\left(\mathrm{2}{x}\right)} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{cos}\:\mathrm{2}{x}−\mathrm{1}\right)+\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}\left(\mathrm{cos}\:\left(\mathrm{4}{x}\right)−\mathrm{1}\right)+\mathrm{2sin}\:^{\mathrm{2}} \left({x}\right)} \\ $$$$\:=\:\frac{\mathrm{3}\left(−\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{2}\left(−\mathrm{2}\right)\left(\mathrm{4}\right)+\mathrm{2}}\:=\frac{−\frac{\mathrm{11}}{\mathrm{2}}}{−\mathrm{14}}\:=\:\frac{\mathrm{11}}{\mathrm{28}} \\ $$

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