Question-93695

Question Number 93695 by i jagooll last updated on 14/May/20

Commented by mathmax by abdo last updated on 14/May/20

f(x) =((6x−^3 (√(27x^3 +2x−1)))/((^3 (√(8x^3 −x)) +x))) ⇒f(x) =((6x−3x(1+(2/(27x^2 ))−(1/(27x^3 )))^(1/3) )/(2x(1−(1/(8x^2 )))^(1/3) +x)) ⇒f(x) ∼((6x−3x(1+(1/3)((2/(27x^2 ))−(1/(27x^3 )))))/(2x(1−(1/(24x^2 )))+x)) ∼((3x)/(3x)) =1 ⇒lim_(x→−∞) f(x)=1

$${f}\left({x}\right)\:=\frac{\mathrm{6}{x}−^{\mathrm{3}} \sqrt{\mathrm{27}{x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{1}}}{\left(^{\mathrm{3}} \sqrt{\mathrm{8}{x}^{\mathrm{3}} −{x}}\:+{x}\right)}\:\Rightarrow{f}\left({x}\right)\:=\frac{\mathrm{6}{x}−\mathrm{3}{x}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{27}{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{27}{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{2}{x}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:+{x}} \\ $$$$\Rightarrow{f}\left({x}\right)\:\sim\frac{\mathrm{6}{x}−\mathrm{3}{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\mathrm{27}{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{27}{x}^{\mathrm{3}} }\right)\right)}{\mathrm{2}{x}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{24}{x}^{\mathrm{2}} }\right)+{x}}\:\sim\frac{\mathrm{3}{x}}{\mathrm{3}{x}}\:=\mathrm{1}\:\Rightarrow{lim}_{{x}\rightarrow−\infty} {f}\left({x}\right)=\mathrm{1} \\ $$

Answered by 1549442205 last updated on 14/May/20

lim_(x→−∞) ((6x−((27x^3 +2x−1))^(1/3) )/( ((8x^3 −x))^(1/3) +x))=lim_(x→−∞) ((6−((27+(2/x^2 )−(1/x^3 )))^(1/3) )/( ((8−(1/x^2 )))^(1/3) +1)) =((6−((27))^(1/3) )/( ((8 ))^(1/3) +1))=((6−3)/(2+1))=1

$$\boldsymbol{\mathrm{li}}\underset{\boldsymbol{\mathrm{x}}\rightarrow−\infty} {\boldsymbol{\mathrm{m}}}\frac{\mathrm{6}\boldsymbol{\mathrm{x}}−\sqrt[{\mathrm{3}}]{\mathrm{27}\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{1}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\boldsymbol{\mathrm{x}}}+\mathrm{x}}=\boldsymbol{\mathrm{li}}\underset{\boldsymbol{\mathrm{x}}\rightarrow−\infty} {\boldsymbol{\mathrm{m}}}\frac{\mathrm{6}−\sqrt[{\mathrm{3}}]{\mathrm{27}+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }}}{\:\sqrt[{\mathrm{3}}]{\mathrm{8}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}+\mathrm{1}} \\ $$$$=\frac{\mathrm{6}−\sqrt[{\mathrm{3}}]{\mathrm{27}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{8}\:\:\:}+\mathrm{1}}=\frac{\mathrm{6}−\mathrm{3}}{\mathrm{2}+\mathrm{1}}=\mathrm{1} \\ $$

Commented by john santu last updated on 14/May/20

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\neq\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\mathrm{sir}\: \\ $$

Commented by john santu last updated on 14/May/20

$$\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{wrong}\: \\ $$

Commented by 1549442205 last updated on 14/May/20

$$\mathrm{excuse}\:\mathrm{me},\mathrm{I}\:\mathrm{wrote}\:\mathrm{fault} \\ $$

Answered by john santu last updated on 14/May/20

lim_(x→−∞) (((6x−x ((27+(2/x^2 )−(1/x^3 )))^(1/(3 )) )/(x ((8−(1/x^2 )))^(1/(3 )) +x))) = lim_(x→−∞) (((−6+3)/(−2−1))) = ((−3)/(−3)) = 1

$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{6}\boldsymbol{{x}}−\boldsymbol{{x}}\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{27}+\frac{\mathrm{2}}{\boldsymbol{{x}}^{\mathrm{2}} }−\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{3}} }}}{\boldsymbol{{x}}\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{8}−\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} }}+\boldsymbol{{x}}}\right)\:= \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\frac{−\mathrm{6}+\mathrm{3}}{−\mathrm{2}−\mathrm{1}}\right)\:=\:\frac{−\mathrm{3}}{−\mathrm{3}}\:=\:\mathrm{1}\: \\ $$

Commented by i jagooll last updated on 14/May/20

cooll man ��

Commented by Ar Brandon last updated on 14/May/20

My man��

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