Question Number 131373 by EDWIN88 last updated on 10/Feb/21
![lim_(x→∞) [ ((x^3 +3x^2 ))^(1/3) +(√(x^2 −2x)) −2x ]=?](https://www.tinkutara.com/question/Q131373.png)
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} }\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}}\:−\mathrm{2}{x}\:\right]=? \\ $$
Answered by aleks041103 last updated on 04/Feb/21
![L=lim_(x→∞) x[ ((x^3 +3x^2 ))^(1/3) +(√(x^2 −2x)) −2x ] ((x^3 +3x^2 ))^(1/3) +(√(x^2 −2x)) −2x= =x[((1+3((1/x))))^(1/3) +(√(1−2((1/x))))−2] ⇒L=lim_(x→∞) ((((1+3((1/x))))^(1/3) +(√(1−2((1/x))))−2)/(((1/x))^2 )) u=(1/x)⇒x→∞⇔u→0^+ L=lim_(u→0) ((((1+3u))^(1/3) +(√(1−2u))−2)/u^2 ) L′Hopital L=lim_(u→0^+ ) (((1+3u)^(−2/3) −(1−2u)^(−1/2) )/(2u))= =lim_(u→0^+ ) (1/2)(((−2)/3)(1+3u)^(−5/3) ×3−((−1)/2)(1−2u)^(−3/2) (−2))= =lim_(u→0^+ ) ((−1)/2)(2(1+3u)^(−5/2) +(1−2u)^(−3/2) )= =−(3/2)](https://www.tinkutara.com/question/Q131379.png)
$${L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\left[\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} }\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}}\:−\mathrm{2}{x}\:\right] \\ $$$$\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} }\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}}\:−\mathrm{2}{x}= \\ $$$$={x}\left[\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{3}\left(\frac{\mathrm{1}}{{x}}\right)}+\sqrt{\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{{x}}\right)}−\mathrm{2}\right] \\ $$$$\Rightarrow{L}=\underset{{x}\rightarrow\infty} {{lim}}\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{3}\left(\frac{\mathrm{1}}{{x}}\right)}+\sqrt{\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{{x}}\right)}−\mathrm{2}}{\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} } \\ $$$${u}=\frac{\mathrm{1}}{{x}}\Rightarrow{x}\rightarrow\infty\Leftrightarrow{u}\rightarrow\mathrm{0}^{+} \\ $$$${L}=\underset{{u}\rightarrow\mathrm{0}} {{lim}}\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{3}{u}}+\sqrt{\mathrm{1}−\mathrm{2}{u}}−\mathrm{2}}{{u}^{\mathrm{2}} } \\ $$$${L}'{Hopital} \\ $$$${L}=\underset{{u}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{\left(\mathrm{1}+\mathrm{3}{u}\right)^{−\mathrm{2}/\mathrm{3}} −\left(\mathrm{1}−\mathrm{2}{u}\right)^{−\mathrm{1}/\mathrm{2}} }{\mathrm{2}{u}}= \\ $$$$=\underset{{u}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{−\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}+\mathrm{3}{u}\right)^{−\mathrm{5}/\mathrm{3}} ×\mathrm{3}−\frac{−\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2}{u}\right)^{−\mathrm{3}/\mathrm{2}} \left(−\mathrm{2}\right)\right)= \\ $$$$=\underset{{u}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{−\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\left(\mathrm{1}+\mathrm{3}{u}\right)^{−\mathrm{5}/\mathrm{2}} +\left(\mathrm{1}−\mathrm{2}{u}\right)^{−\mathrm{3}/\mathrm{2}} \right)= \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by EDWIN88 last updated on 10/Feb/21
![=lim_(x→∞) x((1+(3/x)))^(1/3) +x(√(1−(2/x)))−2x = lim_(x→∞) x [((1+(3/x)))^(1/3) +(√(1−(2/x)))−2 ] = lim_(x→∞) ((((1+(3/x)))^(1/3) +(√(1−(3/x)))−2)/(1/x)) = lim_(x→∞) (((1+(1/x))+(1−(3/(2x)))−2)/(1/x)) = lim_(x→∞) ((−(1/(2x)))/(1/x)) = −(1/2)](https://www.tinkutara.com/question/Q131947.png)
$$\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{x}}}+\mathrm{x}\sqrt{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{x}}}−\mathrm{2x} \\ $$$$\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}x}\:\left[\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{x}}}+\sqrt{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{x}}}−\mathrm{2}\:\right] \\ $$$$\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{x}}}+\sqrt{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{x}}}−\mathrm{2}}{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$$$\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)+\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{2x}}\right)−\mathrm{2}}{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$$$\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{−\frac{\mathrm{1}}{\mathrm{2x}}}{\frac{\mathrm{1}}{\mathrm{x}}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$