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Question Number 115775 by bemath last updated on 28/Sep/20

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt[3]{\frac{(3-\sqrt{x})(\sqrt{x}+2)}{8x-4}}$$

Answered by bobhans last updated on 28/Sep/20

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt[3]{\frac{(3-\sqrt{x})(\sqrt{x}+2)}{8x-4}}=\sqrt[3]{\underset{x\to \mathrm{\infty }}{\lim }\frac{6+\sqrt{x}-x}{8x-4}}$$$$=\sqrt[3]{\underset{x\to \mathrm{\infty }}{\lim }\frac{x(\frac{6}{x}+\frac{1}{\sqrt{x}}-1)}{x(8-\frac{4}{x})}}$$$$=\sqrt[3]{\underset{x\to \mathrm{\infty }}{\lim }\frac{\frac{6}{x}+\frac{1}{\sqrt{x}}-1}{8-\frac{4}{x}}}=-\frac{1}{2}$$

Answered by Dwaipayan Shikari last updated on 28/Sep/20

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt[3]{\frac{(3-\sqrt{\mathrm{x}})(3+\sqrt{\mathrm{x}}-1)}{8\mathrm{x}-4}}=\sqrt[3]{\frac{9-\mathrm{x}-3+\sqrt{\mathrm{x}}}{8\mathrm{x}-4}}$$$$=\sqrt[3]{\frac{\frac{6}{\mathrm{x}}-1+\frac{1}{\sqrt{\mathrm{x}}}}{8-\frac{4}{\mathrm{x}}}}=\sqrt[3]{\frac{-1}{8}}=-\frac{1}{2}$$

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