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Question Number 171552 by 0731619 last updated on 17/Jun/22

Answered by alephzero last updated on 17/Jun/22

$$1-x^3\sim -x^3$$$$\sqrt[3]{-x^3}=-\sqrt[3]{x^3}=-x$$$$\underset{x\to \mathrm{\infty }}{\lim }x+(-x)=0$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }x+\sqrt[3]{1-x^3}=0$$

Answered by Mathematification last updated on 17/Jun/22

$${\mathrm{Lim}}_{\mathrm{x}\to \mathrm{\infty }}(\mathrm{x}+{}^3\sqrt{1-{\mathrm{x}}^3})={\mathrm{Lim}}_{\mathrm{x}\to 0}\left(\frac{1{+}^3\sqrt{{\mathrm{x}}^3-1})}{\mathrm{x}}\right)\stackrel{\mathrm{Lhopital}}{=}\frac{{3\mathrm{x}}^2}{{({\mathrm{x}}^3-1)}^{\frac{2}{3}}}=0$$

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