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 500: 11



Work Step by Step

Given: $\lim\limits_{x\to1}\dfrac{x^3-2x^2+1}{x^3-1}$ This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule. $\lim\limits_{x\to1}\frac{\dfrac{d(x^3-2x^2+1)}{dx}}{\dfrac{d(x^3-1)}{dx}}=\lim\limits_{x\to1}\dfrac{3x^2-4x}{3x^2}$ or, $=\dfrac{3\times1^2-4\times1}{3\times1^2}$ or, $=\dfrac{3-4}{3}$ or,$ =-\dfrac{1}{3}$
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