Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises 2.6 - Page 98: 83



Work Step by Step

\begin{aligned} \lim _{x \rightarrow-\infty}(2 x+\sqrt{4 x^{2}+3 x-2}) &=\lim _{x \rightarrow-\infty}[2 x+\sqrt{4 x^{2}+3 x-2}] \cdot\left[\frac{2 x-\sqrt{4 x^{2}+3 x-2}}{2 x-\sqrt{4 x^{2}+3 x-2}}\right]\\ &=\lim _{x \rightarrow-\infty} \frac{\left(4 x^{2}\right)-\left(4 x^{2}+3 x-2\right)}{2 x-\sqrt{4 x^{2}+3 x-2}} \\ &=\lim _{x \rightarrow-\infty} \frac{-3 x+2}{2 x-\sqrt{4 x^{2}+3 x-2}}\\ &=\lim _{x \rightarrow-\infty} \frac{\frac{-3 x+2}{\sqrt{4^{2}}}}{\frac{-3 x}{\sqrt{x^{2}}}-\sqrt{4+\frac{3}{x^{2}}} x^{2}}\\ &=\lim _{x \rightarrow-\infty} \frac{\frac{-3 x+2}{2 x} \sqrt{4+\frac{3}{x^{2}}-2}}{\frac{2 x}{-x} \sqrt{4+\frac{3}{x^{2}}-x^{2}}} \\ &=\lim _{x \rightarrow-\infty} \frac{3-\frac{2}{x}}{-2-\sqrt{4 \frac{3}{x}}}\\ &=\frac{3-0}{-2-2}\\ &=-\frac{3}{4} \end{aligned}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.