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: 8



Work Step by Step

Divide the numerator and denominator by $x^{2}$, which is the highest power of x in the denominator. Then, we get $\lim\limits_{x \to \infty}\frac{3x^{2}+20x}{4x^{2}+9}=\lim\limits_{x \to \infty}\frac{3-\frac{20}{x}}{4+\frac{9}{x^{2}}}=\frac{3-0}{4+0}=\frac{3}{4}$
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.