Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.7 Limits at Infinity - Exercises - Page 82: 13

Answer

$$-\infty $$

Work Step by Step

\begin{align*} \lim _{x \rightarrow -\infty}\frac{7x^2-9}{4x+3}&= \lim _{x \rightarrow -\infty}\frac{x^2}{x}\frac{7-9x^{-2}}{4+3x^{-1}}\\ &=\lim _{x \rightarrow -\infty}\frac{x^2}{x}\lim _{x \rightarrow -\infty}\frac{7-9x^{-2}}{4+3x^{-1}}\\ &=\frac{7}{4}\lim _{x \rightarrow -\infty} x \\ &=-\infty.\\ \end{align*}
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.