Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.4 Limits at Infinity; Horizontal Asymptotes - 3.4 Exercises - Page 242: 22


$$\frac{-3}{4} $$

Work Step by Step

Given $$ \lim _{x \rightarrow -\infty}(\sqrt{4 x^{2}+3x}+2 x)$$ Then \begin{aligned} \lim _{x \rightarrow-\infty}(\sqrt{4 x^{2}+3x}+2 x) &=\lim _{x \rightarrow -\infty} \frac{(\sqrt{4 x^{2}+3x}+2 x)(\sqrt{4 x^{2}+3x}-2 x)}{ \sqrt{4 x^{2}+3x}-2 x}\\ &=\lim _{x \rightarrow -\infty} \frac{(\sqrt{4 x^{2}+3x})^2-(2 x)^2 }{ \sqrt{4 x^{2}+3x}-2 x} \\ &=\lim _{x \rightarrow -\infty} \frac{4 x^{2}+3x-4 x^2 }{ \sqrt{4 x^{2}+3x}-2 x}\\ &=\lim _{x \rightarrow- \infty} \frac{3x}{ \sqrt{4 x^{2}+3x}-2 x} \cdot \frac{1 / x}{1 / x} \\ &=\lim _{x \rightarrow -\infty} \frac{3x / x}{\sqrt{4 x^{2} / x^{2}+3x / x^{2}}-2}\\ &=\lim _{x \rightarrow- \infty} \frac{3}{\sqrt{4+3 / x}-2}\\ &=\frac{3}{-2-2}=\frac{-3}{4} \end{aligned}
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