University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 248: 7

Answer

$\dfrac{1}{4}$

Work Step by Step

Consider: $\lim\limits_{x \to 2}f(x)=\lim\limits_{x \to 2}\dfrac{x -2}{x^2-4}$ Now, $f(2)=\dfrac{2 -2}{4-4}=\dfrac{0}{0}$ The limit shows an indeterminate form. Thus, apply L-Hospital's rule: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ Thus: $\lim\limits_{x \to 2}\dfrac{1}{2x}=\dfrac{1}{2(2)}=\dfrac{1}{4}$
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