Answer
$\lim\limits_{x \to 2^+}\frac{x^2-2x-8}{x^2-5x+6}=\infty$
Work Step by Step
$\lim\limits_{x \to 2^+}\frac{x^2-2x-8}{x^2-5x+6}=\lim\limits_{x \to 2^+}\frac{x^2-2x-8}{(x-2)(x-3)}=\frac{(-8)^+}{0^+\times(-1^+)}=\frac{(-8)^+}{0^-}=\infty$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
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 - Page 94: 41
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