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Question Number 110861 by apriani last updated on 31/Aug/20

Answered by bemath last updated on 31/Aug/20

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{{\mathrm{x}}^3-{6\mathrm{x}}^2}-(\mathrm{x}+2)=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{({\mathrm{x}}^3-{6\mathrm{x}}^2)-({\mathrm{x}}^2+4\mathrm{x}+4)}{\sqrt{{\mathrm{x}}^3-{6\mathrm{x}}^2}+(\mathrm{x}+2)}=$$$$\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^3-{7\mathrm{x}}^2-4\mathrm{x}-4}{\sqrt{{\mathrm{x}}^3-{6\mathrm{x}}^2}+(\mathrm{x}+2)}=$$$$\underset{x\to 0}{\lim }\frac{1-\frac{7}{\mathrm{x}}-\frac{4}{{\mathrm{x}}^2}-\frac{4}{{\mathrm{x}}^3}}{\sqrt{\frac{1}{{\mathrm{x}}^3}-\frac{6}{{\mathrm{x}}^4}}+(\frac{1}{{\mathrm{x}}^2}+\frac{2}{{\mathrm{x}}^3})}=$$$$\frac{1-0}{0}=\mathrm{\infty }$$$$$$

Answered by Aziztisffola last updated on 31/Aug/20

$$\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\sqrt{{\mathrm{x}}^3-{6\mathrm{x}}^2}-\mathrm{x}-2$$$$\underset{\mathrm{x}\to \mathrm{\infty }}{\lim x}\sqrt{\mathrm{x}-6}-\mathrm{x}-2$$$$\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\mathrm{x}(\sqrt{\mathrm{x}-6}-1-\frac{2}{\mathrm{x}})=\mathrm{\infty }\times \mathrm{\infty }=\mathrm{\infty }$$

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