Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.2 - The Limit of a Function - 2.2 Exercises: 41

Answer

$\lim\limits_{x \to 2^+}\frac{x^2-2x-8}{x^2-5x+6}=\infty$

Work Step by Step

$\lim\limits_{x \to 2^+}\frac{x^2-2x-8}{x^2-5x+6}=\lim\limits_{x \to 2^+}\frac{x^2-2x-8}{(x-2)(x-3)}=\frac{(-8)^+}{0^+\times(-1^+)}=\frac{(-8)^+}{0^-}=\infty$
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