Question Number 186066 by TUN last updated on 31/Jan/23
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\left(\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{7}}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{2}}−{x}+\mathrm{1}\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}−\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} }\right) \\ $$$$ \\ $$
Answered by a.lgnaoui last updated on 31/Jan/23
![x((√(4x^2 −12x+7)) −(√(x^2 −4x−2)) −x+1) =2x(√(x^2 −3x+(7/4))) −x(√(x^2 −4x−2)) −x^2 +x lim_(x→∞) [2x(√(x^2 −3x+(7/4))) −x(√(x^2 −4x−2)) −(x^2 −x)] =lim_(x→∞) [x(((2(4x−3x+((15)/4)))/( (√(x^2 −3x+(7/4))) +(√(x^2 −4x−2)) )))−x(x−1)] [(((2x^2 )/(x(√(1−(3/x)+(7/(4x^2 )))) +x(√(1−(4/x)−(2/x^2 ))))))−x^2 (1−(1/x))] =lim_(x→∞) x−x^2 (1−(1/x))=lim_(x→∞) (−x^2 )=−∞](https://www.tinkutara.com/question/Q186093.png)
$$\:{x}\left(\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{7}}\:−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{2}}\:\:−{x}+\mathrm{1}\right) \\ $$$$=\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\frac{\mathrm{7}}{\mathrm{4}}}\:−{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{2}}\:\:−{x}^{\mathrm{2}} +{x} \\ $$$$\mathrm{lim}_{{x}\rightarrow\infty} \left[\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\frac{\mathrm{7}}{\mathrm{4}}}\:−{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{2}}\:−\left({x}^{\mathrm{2}} −{x}\right)\right] \\ $$$$=\mathrm{lim}_{{x}\rightarrow\infty} \left[{x}\left(\frac{\mathrm{2}\left(\mathrm{4}{x}−\mathrm{3}{x}+\frac{\mathrm{15}}{\mathrm{4}}\right)}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\frac{\mathrm{7}}{\mathrm{4}}}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{2}}\:}\right)−{x}\left({x}−\mathrm{1}\right)\right] \\ $$$$\left[\left(\frac{\mathrm{2}{x}^{\mathrm{2}} }{{x}\sqrt{\mathrm{1}−\frac{\mathrm{3}}{{x}}+\frac{\mathrm{7}}{\mathrm{4}{x}^{\mathrm{2}} }}\:+{x}\sqrt{\mathrm{1}−\frac{\mathrm{4}}{{x}}−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }}}\right)−{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\right] \\ $$$$=\mathrm{lim}_{{x}\rightarrow\infty} {x}−{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=\mathrm{lim}_{{x}\rightarrow\infty} \left(−{x}^{\mathrm{2}} \right)=−\infty \\ $$
Commented by Frix last updated on 01/Feb/23
$$\mathrm{Wrong}. \\ $$
Answered by Frix last updated on 01/Feb/23
![lim_(x→∞) x((√(4x^2 −12x+7))−(√(x^2 −4x−2))−x+1) = lim_(t→0^+ ) ((t−1+(√(7t^2 −12t+4))−(√(−2t^2 +4t+1)))/t^2 ) = =lim_(t→0^+ ) (((d^2 [t−1+(√(7t^2 −12t+4))−(√(−2t^2 +4t+1))])/dt^2 )/((d^2 [t^2 ])/dt^2 )) = =lim_(t→0^+ ) (((6/((−2t^2 −4t+1)^(3/2) ))−(8/((7t^2 −12t+4)^(3/2) )))/2) = =(5/2) Similar procedure for the 2^(nd) one leads to (5/6)](https://www.tinkutara.com/question/Q186095.png)
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\left(\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{7}}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{2}}−{x}+\mathrm{1}\right)\:= \\ $$$$\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{{t}−\mathrm{1}+\sqrt{\mathrm{7}{t}^{\mathrm{2}} −\mathrm{12}{t}+\mathrm{4}}−\sqrt{−\mathrm{2}{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{1}}}{{t}^{\mathrm{2}} }\:= \\ $$$$=\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\frac{{d}^{\mathrm{2}} \left[{t}−\mathrm{1}+\sqrt{\mathrm{7}{t}^{\mathrm{2}} −\mathrm{12}{t}+\mathrm{4}}−\sqrt{−\mathrm{2}{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{1}}\right]}{{dt}^{\mathrm{2}} }}{\frac{{d}^{\mathrm{2}} \left[{t}^{\mathrm{2}} \right]}{{dt}^{\mathrm{2}} }}\:= \\ $$$$=\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\frac{\mathrm{6}}{\left(−\mathrm{2}{t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }−\frac{\mathrm{8}}{\left(\mathrm{7}{t}^{\mathrm{2}} −\mathrm{12}{t}+\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }}{\mathrm{2}}\:= \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{Similar}\:\mathrm{procedure}\:\mathrm{for}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{one}\:\mathrm{leads}\:\mathrm{to}\:\frac{\mathrm{5}}{\mathrm{6}} \\ $$
Answered by cortano1 last updated on 01/Feb/23
![Without L′Hopital L=lim_(x→∞) x[(√(4x^2 −12x+7))−(√(x^2 −4x−2))−x+1 ] = lim_(x→∞) x[ (√(4x^2 −12x+7))−(2x−3)+(x−2)−(√(x^2 −4x−2)) ] = lim_(x→∞) x[((−2)/( (√(4x^2 −12x+7))+(2x−3))) + (6/(x−2+(√(x^2 −4x−2)))) ] = ((−2)/( (√4)+2)) +(6/(1+1)) = (5/2)](https://www.tinkutara.com/question/Q186104.png)
$${Without}\:{L}'{Hopital}\: \\ $$$$\:{L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}\left[\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{7}}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{2}}−{x}+\mathrm{1}\:\right] \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}\left[\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{7}}−\left(\mathrm{2}{x}−\mathrm{3}\right)+\left({x}−\mathrm{2}\right)−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{2}}\:\right] \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}\left[\frac{−\mathrm{2}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{7}}+\left(\mathrm{2}{x}−\mathrm{3}\right)}\:+\:\frac{\mathrm{6}}{{x}−\mathrm{2}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{2}}}\:\right] \\ $$$$=\:\frac{−\mathrm{2}}{\:\sqrt{\mathrm{4}}+\mathrm{2}}\:+\frac{\mathrm{6}}{\mathrm{1}+\mathrm{1}}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$