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Question Number 65473 by Tawa1 last updated on 30/Jul/19

Commented by mathmax by abdo last updated on 30/Jul/19

let A(x) =(√((x^3 −1)/(x+1)))−^3 (√((x^5 +1)/(x^2 −2x))) ⇒for x>0 A(x) =(√((x^3 (1−(1/x^3 )))/(x(1+(1/x)))))−^3 (√((x^5 (1+(1/x^5 )))/(x^2 (1−(2/x))))) =x(√((1−(1/x^3 ))/(1+(1/x))))−x(^3 (√((1+(1/x^5 ))/(1−(2/x))))) (1/(1+(1/x))) ∼1−(1/x) ⇒((1−(1/x^3 ))/(1+(1/x))) ∼(1−(1/x^3 ))(1+(1/x))=1+(1/x)−(1/x^3 )−(1/x^4 ) ⇒ (√((1−(1/x^3 ))/(1+(1/x)))) ∼1+(1/(2x))−(1/(2x^3 ))−(1/(2x^4 )) (1/(1−(2/x)))∼1+(2/x) ⇒((1+(1/x^5 ))/(1−(2/x))) ∼(1+(1/x^5 ))(1−(2/x))=1−(2/x)+(1/x^5 )−(2/x^6 ) and^3 (√((1+(1/x^5 ))/(1−(2/x)))) ∼(1−(2/x) +(1/x^5 )−(2/x^6 ))^(1/3) ∼1+(1/3)(−(2/x)+(1/x^5 )−(2/x^6 )) =1−(2/(3x)) +(1/(3x^5 )) −(2/(3x^6 )) ⇒ A(x) ∼x(1 +(1/(2x))−(1/(2x^3 ))−(1/(2x^4 )))−x(1−(2/(3x)) +(1/(3x^5 ))−(2/(3x^6 ))) =(1/2)−(1/(2x^2 ))−(1/(2x^3 )) +(2/3)−(1/(3x^4 )) +(2/(3x^5 )) ⇒lim_(x→+∞) A(x) =(7/6)

$${let}\:\:{A}\left({x}\right)\:=\sqrt{\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}+\mathrm{1}}}−^{\mathrm{3}} \sqrt{\frac{{x}^{\mathrm{5}} +\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}{x}}}\:\Rightarrow{for}\:{x}>\mathrm{0} \\ $$$${A}\left({x}\right)\:=\sqrt{\frac{{x}^{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)}{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)}}−\:^{\mathrm{3}} \sqrt{\frac{{x}^{\mathrm{5}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{2}}{{x}}\right)}} \\ $$$$={x}\sqrt{\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}}−{x}\left(\:^{\mathrm{3}} \sqrt{\frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}{\mathrm{1}−\frac{\mathrm{2}}{{x}}}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}\:\sim\mathrm{1}−\frac{\mathrm{1}}{{x}}\:\Rightarrow\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}\:\sim\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}+\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:\Rightarrow \\ $$$$\sqrt{\frac{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}}\:\sim\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{4}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{2}}{{x}}}\sim\mathrm{1}+\frac{\mathrm{2}}{{x}}\:\Rightarrow\frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}{\mathrm{1}−\frac{\mathrm{2}}{{x}}}\:\sim\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)\left(\mathrm{1}−\frac{\mathrm{2}}{{x}}\right)=\mathrm{1}−\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }−\frac{\mathrm{2}}{{x}^{\mathrm{6}} } \\ $$$${and}\:^{\mathrm{3}} \sqrt{\frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}{\mathrm{1}−\frac{\mathrm{2}}{{x}}}}\:\sim\left(\mathrm{1}−\frac{\mathrm{2}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }−\frac{\mathrm{2}}{{x}^{\mathrm{6}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \sim\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\left(−\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{5}} }−\frac{\mathrm{2}}{{x}^{\mathrm{6}} }\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}{x}}\:+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{5}} }\:−\frac{\mathrm{2}}{\mathrm{3}{x}^{\mathrm{6}} }\:\Rightarrow \\ $$$${A}\left({x}\right)\:\sim{x}\left(\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{4}} }\right)−{x}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}{x}}\:+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{5}} }−\frac{\mathrm{2}}{\mathrm{3}{x}^{\mathrm{6}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{3}} }\:+\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{4}} }\:+\frac{\mathrm{2}}{\mathrm{3}{x}^{\mathrm{5}} }\:\Rightarrow{lim}_{{x}\rightarrow+\infty} \:{A}\left({x}\right)\:=\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$ \\ $$

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