Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.4 - The Precise Definition of a Limit - 2.4 Exercises - Page 114: 21

Answer

$\lim\limits_{x \to 4} \frac{x^2-2x-8}{x-4} = 6$

Work Step by Step

Let $\epsilon \gt 0$ be given. Let $\delta = \epsilon$ Suppose that $\vert x-4 \vert \lt \delta$. Then: $\vert \frac{x^2-2x-8}{x-4}-6 \vert = \vert \frac{(x-4)(x+2)}{x-4}-6 \vert = \vert (x+2)-6 \vert = \vert x-4 \vert \lt \delta = \epsilon$ Therefore, $\lim\limits_{x \to 4} \frac{x^2-2x-8}{x-4} = 6$
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