Question Number 190082 by Rupesh123 last updated on 27/Mar/23
Answered by a.lgnaoui last updated on 29/Mar/23
![L=(e^x^3 /x^3 )−((3(√((1−x)(1+x^2 +x))))/((x^3 −1)+1))+((√((1−x)(1+x)))/((x^2 −1)+1)) (e^x /x^3 )−((3(√(1−x)))/x^3 )+((√(1−x))/x^2 ) =(e^x^3 /x^3 )−[((3(√(1−x)))/x^3 )−x((√((1−x)))/x^3 ))] =(1/x^3 )(e^x^3 −((3(√(1−x)))/1)) x=(1/X) x→0 X→+∞ X^3 (e^(1/X^3 ) −3(√(1−(1/X))) ) lim_(X→∞) L=+∞(1−3)=−∞ β=−∞ 3!β=−∞](https://www.tinkutara.com/question/Q190158.png)
$${L}=\frac{{e}^{{x}^{\mathrm{3}} } }{{x}^{\mathrm{3}} }−\frac{\mathrm{3}\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} +{x}\right)}}{\left({x}^{\mathrm{3}} −\mathrm{1}\right)+\mathrm{1}}+\frac{\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}} \\ $$$$\frac{{e}^{{x}} }{{x}^{\mathrm{3}} }−\frac{\mathrm{3}\sqrt{\mathrm{1}−{x}}}{{x}^{\mathrm{3}} }+\frac{\sqrt{\mathrm{1}−{x}}}{{x}^{\mathrm{2}} } \\ $$$$\left.=\frac{{e}^{{x}^{\mathrm{3}} } }{{x}^{\mathrm{3}} }−\left[\frac{\mathrm{3}\sqrt{\mathrm{1}−{x}}}{{x}^{\mathrm{3}} }−{x}\frac{\sqrt{\left(\mathrm{1}−{x}\right)}}{{x}^{\mathrm{3}} }\right)\right] \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\left({e}^{{x}^{\mathrm{3}} } −\frac{\mathrm{3}\sqrt{\mathrm{1}−{x}}}{\mathrm{1}}\right) \\ $$$${x}=\frac{\mathrm{1}}{{X}}\:\:\:{x}\rightarrow\mathrm{0}\:\:\:\:{X}\rightarrow+\infty \\ $$$${X}^{\mathrm{3}} \left({e}^{\frac{\mathrm{1}}{{X}^{\mathrm{3}} }} −\mathrm{3}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{X}}}\:\right) \\ $$$${lim}_{{X}\rightarrow\infty} {L}=+\infty\left(\mathrm{1}−\mathrm{3}\right)=−\infty \\ $$$$\:\beta=−\infty \\ $$$$\mathrm{3}!\beta=−\infty \\ $$