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Question Number 146263 by bemath last updated on 12/Jul/21

$$\underset{x\to -\mathrm{\infty }}{\lim }(-2x+1-\sqrt{4x^2-5x+8})=?$$

Answered by gsk2684 last updated on 12/Jul/21

$$\underset{x\to -\mathrm{\infty }}{\lim }\frac{{(-2x+1)}^2-(4x^2-5x+8)}{-2x+1+\sqrt{4x^2-5x+8}}$$$$\underset{x\to -\mathrm{\infty }}{\lim }\frac{x-7}{-2x+1-x\sqrt{4-\frac{5}{x}+\frac{8}{x^2}}}$$$$\frac{1}{-2-\sqrt{4}}=-\frac{1}{4}$$

Commented by mr W last updated on 12/Jul/21

$$correct$$

Answered by iloveisrael last updated on 13/Jul/21

$$\underset{x\to -\mathrm{\infty }}{\lim }(-2\mathrm{x}+1-\sqrt{{4\mathrm{x}}^2-5\mathrm{x}+8})=?$$$$\mathrm{solution}:$$$$\underset{x\to -\mathrm{\infty }}{\lim }(-2\mathrm{x}+1-\sqrt{{4\mathrm{x}}^2-5\mathrm{x}+8})$$$$=\underset{x\to -\mathrm{\infty }}{\lim }(-2\mathrm{x}+1-\sqrt{4}\mid \mathrm{x}-\frac{5}{2.4}\mid )$$$$=\underset{x\to -\mathrm{\infty }}{\lim }(-2\mathrm{x}+1-2(-\mathrm{x}+\frac{5}{8}))$$$$=\underset{x\to -\mathrm{\infty }}{\lim }(-2\mathrm{x}+1+2\mathrm{x}-\frac{5}{4})$$$$=\underset{x\to -\mathrm{\infty }}{\lim }(1-\frac{5}{4})=-\frac{1}{4}.✓$$

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