Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.8 Indeterminate Forms and I'Hospital's Rule - 6.8 Exercises - Page 499: 5


$ 2.25$

Work Step by Step

From the given graph, we have $\lim\limits_{x \to 2} \dfrac{f(x)}{g(x)} =\lim\limits_{x \to 2} \dfrac{f'(x)}{g'(x)} $ or, $\lim\limits_{x \to 2}\dfrac{1.8(x-2)}{\dfrac{4}{5}(x-2)} = \dfrac{0}{0}$ This shows is an indeterminate form. Thus, on applying L'Hospital, we get $=\lim\limits_{x \to 2} \dfrac{1.8(1)}{\dfrac{4}{5}(1)}$ $ = 2.25$
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