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Question Number 111923 by bemath last updated on 05/Sep/20
(√(bemath )) lim_(x→0) ((((cos 4x))^(1/(3 )) −1)/(cos 3x−cos 9x)) ?
$$\sqrt{{bemath}\:} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{4}{x}}\:−\mathrm{1}}{\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{9}{x}}\:? \\ $$
Answered by bobhans last updated on 05/Sep/20
lim_(x→0) ((((cos 4x))^(1/(3 )) −1)/(cos 3x − cos 9x)) ? recall a^3 −b^3 =(a−b)(a^2 +b^2 +ab) then ((cos 4x))^(1/(3 )) −1 = ((cos 4x−1)/( ((cos^2 4x))^(1/(3 )) +1+((cos 4x))^(1/(3 )) )) lim_(x→0) ((cos 4x−1)/(cos 3x−cos 9x)) .lim_(x→0) (1/( ((cos^2 4x))^(1/(3 )) +1 +((cos 4x))^(1/(3 )) )) = lim_(x→0) ((((−1)/2).16x^2 )/(−(1/2).9x^2 +(1/2).81x^2 )) × (1/3) = −(1/3).lim_(x→0) ((16x^2 )/(72x^2 )) = −(2/(27))
$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{4x}}\:−\mathrm{1}}{\mathrm{cos}\:\mathrm{3x}\:−\:\mathrm{cos}\:\mathrm{9x}}\:? \\ $$$$\mathrm{recall}\:\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} =\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{ab}\right) \\ $$$$\mathrm{then}\:\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{4x}}\:−\mathrm{1}\:=\:\frac{\mathrm{cos}\:\mathrm{4x}−\mathrm{1}}{\:\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:^{\mathrm{2}} \mathrm{4x}}+\mathrm{1}+\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{4x}}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{4x}−\mathrm{1}}{\mathrm{cos}\:\mathrm{3x}−\mathrm{cos}\:\mathrm{9x}}\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:^{\mathrm{2}} \mathrm{4x}}\:+\mathrm{1}\:+\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{4x}}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{−\mathrm{1}}{\mathrm{2}}.\mathrm{16x}^{\mathrm{2}} }{−\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{9x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{81x}^{\mathrm{2}} }\:×\:\frac{\mathrm{1}}{\mathrm{3}}\:= \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{16x}^{\mathrm{2}} }{\mathrm{72x}^{\mathrm{2}} }\:=\:−\frac{\mathrm{2}}{\mathrm{27}}\: \\ $$
Answered by mathmax by abdo last updated on 05/Sep/20
let f(x) =(((^3 (√(cos(4x))))−1)/(cos(3x)−cos(9x))) ⇒f(x) =(((cos(4x)^(1/3) −1)/(cos(3x)−cos(9x))) we hsve cos(4x) ∼1−(((4x)^2 )/2) =1−8x^2 (x∈v(0)) ⇒ (cos(4x))^(1/3) ∼(1−8x^2 )^(1/3) ∼1−(8/3)x^2 also cos(3x)∼1−((9x^2 )/2) and cos(9x)∼1−((81x^2 )/2) ⇒f(x)∼((−(8/3)x^2 )/(−((9x^2 )/2)+((81)/2)x^2 )) ⇒ f(x)∼−(8/(3.36)) =−(2/(27)) ⇒lim_(x→0) f(x) =−(2/(27))
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\left(^{\mathrm{3}} \sqrt{\mathrm{cos}\left(\mathrm{4x}\right)}\right)−\mathrm{1}}{\mathrm{cos}\left(\mathrm{3x}\right)−\mathrm{cos}\left(\mathrm{9x}\right)}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\left(\mathrm{cos}\left(\mathrm{4x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1}\right.}{\mathrm{cos}\left(\mathrm{3x}\right)−\mathrm{cos}\left(\mathrm{9x}\right)} \\ $$$$\mathrm{we}\:\mathrm{hsve}\:\mathrm{cos}\left(\mathrm{4x}\right)\:\sim\mathrm{1}−\frac{\left(\mathrm{4x}\right)^{\mathrm{2}} }{\mathrm{2}}\:\:=\mathrm{1}−\mathrm{8x}^{\mathrm{2}} \:\:\:\left(\mathrm{x}\in\mathrm{v}\left(\mathrm{0}\right)\right)\:\Rightarrow \\ $$$$\left(\mathrm{cos}\left(\mathrm{4x}\right)\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\sim\left(\mathrm{1}−\mathrm{8x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\sim\mathrm{1}−\frac{\mathrm{8}}{\mathrm{3}}\mathrm{x}^{\mathrm{2}} \:\mathrm{also}\:\mathrm{cos}\left(\mathrm{3x}\right)\sim\mathrm{1}−\frac{\mathrm{9x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{cos}\left(\mathrm{9x}\right)\sim\mathrm{1}−\frac{\mathrm{81x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\frac{−\frac{\mathrm{8}}{\mathrm{3}}\mathrm{x}^{\mathrm{2}} }{−\frac{\mathrm{9x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{81}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim−\frac{\mathrm{8}}{\mathrm{3}.\mathrm{36}}\:=−\frac{\mathrm{2}}{\mathrm{27}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)\:=−\frac{\mathrm{2}}{\mathrm{27}} \\ $$

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