Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises: 15

Answer

$\lim\limits_{x\to\infty}\frac{3x-2}{2x+1}=\frac{3}{2}$

Work Step by Step

$$A=\lim\limits_{x\to\infty}\frac{3x-2}{2x+1}$$ Divide both numerator and denominator by $x$ $A=\lim\limits_{x\to\infty}\frac{\frac{3x-2}{x}}{\frac{2x+1}{x}}$ $A=\lim\limits_{x\to\infty}\frac{3-\frac{2}{x}}{2+\frac{1}{x}}$ $A=\frac{\lim\limits_{x\to\infty}(3-\frac{2}{x})}{\lim\limits_{x\to\infty}(2+\frac{1}{x})}$ $A=\frac{\lim\limits_{x\to\infty}3-\lim\limits_{x\to\infty}\frac{2}{x}}{\lim\limits_{x\to\infty}2+\lim\limits_{x\to\infty}\frac{1}{x}}$ $A=\frac{3-0}{2+0}$ (Theorem 5) $A=\frac{3}{2}$
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.