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: 10



Work Step by Step

Given: $\lim\limits_{x\to-2}\dfrac{x^3+8}{x+2}$ This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule. $\lim_{x\to-2}\dfrac{\dfrac{d(x^3+8)}{dx}}{\dfrac{d(x+2)}{dx}}=\lim\limits_{x\to-2}\dfrac{3x^2}{1}$ or, $=\lim\limits_{x\to-2}(3x^2)$ or, $=3\times(-2)^2$ or, $=3\times4$ or, $=12$
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