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 - Page 69: 24

Answer

$-\infty$

Work Step by Step

$$\lim\limits_{x \to 4^+} \frac{3-x}{x^2 -2x - 8} = \lim\limits_{x \to 4^+} (3-x) \cdot \lim\limits_{x \to 4^+} \frac{1}{x^2 -2x - 8} = -1 \cdot \infty = -\infty $$
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