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Question Number 131373 by EDWIN88 last updated on 10/Feb/21

$$\underset{x\to \mathrm{\infty }}{\lim }[\sqrt[3]{x^3+3x^2}+\sqrt{x^2-2x}-2x]=?$$

Answered by aleks041103 last updated on 04/Feb/21

$$L=\underset{x\to \mathrm{\infty }}{\lim }x[\sqrt[3]{x^3+3x^2}+\sqrt{x^2-2x}-2x]$$$$\sqrt[3]{x^3+3x^2}+\sqrt{x^2-2x}-2x=$$$$=x[\sqrt[3]{1+3\left(\frac{1}{x}\right)}+\sqrt{1-2\left(\frac{1}{x}\right)}-2]$$$$\Rightarrow L=\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt[3]{1+3\left(\frac{1}{x}\right)}+\sqrt{1-2\left(\frac{1}{x}\right)}-2}{{\left(\frac{1}{x}\right)}^2}$$$$u=\frac{1}{x}\Rightarrow x\to \mathrm{\infty }\iff u\to 0^{+}$$$$L=\underset{u\to 0}{lim}\frac{\sqrt[3]{1+3u}+\sqrt{1-2u}-2}{u^2}$$$$L^{\prime }Hopital$$$$L=\underset{u\to 0^{+}}{lim}\frac{{(1+3u)}^{-2/3}-{(1-2u)}^{-1/2}}{2u}=$$$$=\underset{u\to 0^{+}}{lim}\frac{1}{2}(\frac{-2}{3}{(1+3u)}^{-5/3}\times 3-\frac{-1}{2}{(1-2u)}^{-3/2}(-2))=$$$$=\underset{u\to 0^{+}}{lim}\frac{-1}{2}(2{(1+3u)}^{-5/2}+{(1-2u)}^{-3/2})=$$$$=-\frac{3}{2}$$

Answered by EDWIN88 last updated on 10/Feb/21

$$=\underset{x\to \mathrm{\infty }}{\lim }\mathrm{x}\sqrt[3]{1+\frac{3}{\mathrm{x}}}+\mathrm{x}\sqrt{1-\frac{2}{\mathrm{x}}}-2\mathrm{x}$$$$=\underset{x\to \mathrm{\infty }}{\lim x}[\sqrt[3]{1+\frac{3}{\mathrm{x}}}+\sqrt{1-\frac{2}{\mathrm{x}}}-2]$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt[3]{1+\frac{3}{\mathrm{x}}}+\sqrt{1-\frac{3}{\mathrm{x}}}-2}{\frac{1}{\mathrm{x}}}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{(1+\frac{1}{\mathrm{x}})+(1-\frac{3}{2\mathrm{x}})-2}{\frac{1}{\mathrm{x}}}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{-\frac{1}{2\mathrm{x}}}{\frac{1}{\mathrm{x}}}=-\frac{1}{2}$$

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