Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises: 28

Answer

$$\lim\limits_{x\to-\infty}(\sqrt{4x^2+3x}+2x)=-\frac{3}{4}$$

Work Step by Step

$$A=\lim\limits_{x\to-\infty}(\sqrt{4x^2+3x}+2x)$$$$A=\lim\limits_{x\to-\infty}\Bigg[(\sqrt{4x^2+3x}+2x)\times\frac{\sqrt{4x^2+3x}-2x}{\sqrt{4x^2+3x}-2x}\Bigg]$$$$A=\lim\limits_{x\to-\infty}\frac{(4x^2+3x)-4x^2}{\sqrt{4x^2+3x}-2x}$$$$A=\lim\limits_{x\to-\infty}\frac{3x}{\sqrt{4x^2+3x}-2x}$$$$A=\lim\limits_{x\to-\infty}\frac{X}{Y}$$ Divide both numerator and denominator by $x$ Notice that as $x\to-\infty$, we have $x\lt0$. Therefore, $\sqrt{x^2}=|x|=-x$ - The numerator: $X=\frac{3x}{x}=3$ - The denominator: $Y=\frac{\sqrt{4x^2+3x}-2x}{x}=\frac{\sqrt{4x^2+3x}}{x}-\frac{2x}{x}=\frac{\sqrt{4x^2+3x}}{-\sqrt{x^2}}-2=-\sqrt{4+\frac{3}{x}}-2$ Therefore, $$A=\lim\limits_{x\to-\infty}\frac{3}{-\sqrt{4+\frac{3}{x}}-2}$$$$A=\frac{3}{-\sqrt{4+\lim\limits_{x\to-\infty}(\frac{3}{x})}-2}$$$$A=\frac{3}{-\sqrt{4+0}-2}$$$$A=-\frac{3}{4}$$

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