Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 1 - Limits and Continuity - 1.2 Computing Limits - Exercises Set 1.2: 12

Answer

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

Work Step by Step

3x² - x - 2 = (x - 1)(3x + 2) 2x² + x - 3 = (x - 1)(2x + 3), so: $\frac{3x² - x - 2}{2x² + x - 3}$ = $\frac{(x-1)(3x+2)}{(x-1)(2x+3)}$ = $\frac{3x+2}{2x+3}$ $\lim\limits_{x \to 1}\frac{3x² - x - 2}{2x² + x - 3}$ = $\lim\limits_{x \to 1}\frac{3x+2}{2x+3}$ = $\frac{3\times1+2}{2\times1+3}$ = 1
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