Question Number 93695 by i jagooll last updated on 14/May/20
Commented by mathmax by abdo last updated on 14/May/20
$${f}\left({x}\right)\:=\frac{\mathrm{6}{x}−^{\mathrm{3}} \sqrt{\mathrm{27}{x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{1}}}{\left(^{\mathrm{3}} \sqrt{\mathrm{8}{x}^{\mathrm{3}} −{x}}\:+{x}\right)}\:\Rightarrow{f}\left({x}\right)\:=\frac{\mathrm{6}{x}−\mathrm{3}{x}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{27}{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{27}{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{2}{x}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:+{x}} \\ $$$$\Rightarrow{f}\left({x}\right)\:\sim\frac{\mathrm{6}{x}−\mathrm{3}{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{2}}{\mathrm{27}{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{27}{x}^{\mathrm{3}} }\right)\right)}{\mathrm{2}{x}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{24}{x}^{\mathrm{2}} }\right)+{x}}\:\sim\frac{\mathrm{3}{x}}{\mathrm{3}{x}}\:=\mathrm{1}\:\Rightarrow{lim}_{{x}\rightarrow−\infty} {f}\left({x}\right)=\mathrm{1} \\ $$
Answered by 1549442205 last updated on 14/May/20
$$\boldsymbol{\mathrm{li}}\underset{\boldsymbol{\mathrm{x}}\rightarrow−\infty} {\boldsymbol{\mathrm{m}}}\frac{\mathrm{6}\boldsymbol{\mathrm{x}}−\sqrt[{\mathrm{3}}]{\mathrm{27}\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{1}}}{\sqrt[{\mathrm{3}}]{\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\boldsymbol{\mathrm{x}}}+\mathrm{x}}=\boldsymbol{\mathrm{li}}\underset{\boldsymbol{\mathrm{x}}\rightarrow−\infty} {\boldsymbol{\mathrm{m}}}\frac{\mathrm{6}−\sqrt[{\mathrm{3}}]{\mathrm{27}+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }}}{\sqrt[{\mathrm{3}}]{\mathrm{8}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}+\mathrm{1}} \\ $$$$=\frac{\mathrm{6}−\sqrt[{\mathrm{3}}]{\mathrm{27}}}{\sqrt[{\mathrm{3}}]{\mathrm{8}\:\:\:}+\mathrm{1}}=\frac{\mathrm{6}−\mathrm{3}}{\mathrm{2}+\mathrm{1}}=\mathrm{1} \\ $$
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
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{6}\boldsymbol{{x}}−\boldsymbol{{x}}\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{27}+\frac{\mathrm{2}}{\boldsymbol{{x}}^{\mathrm{2}} }−\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{3}} }}}{\boldsymbol{{x}}\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{8}−\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} }}+\boldsymbol{{x}}}\right)\:= \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\frac{−\mathrm{6}+\mathrm{3}}{−\mathrm{2}−\mathrm{1}}\right)\:=\:\frac{−\mathrm{3}}{−\mathrm{3}}\:=\:\mathrm{1}\: \\ $$
Commented by i jagooll last updated on 14/May/20
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Commented by Ar Brandon last updated on 14/May/20
My man��