Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - Chapter Review Exercises - Page 94: 17

Answer

$2$

Work Step by Step

\begin{align*} \lim _{x \rightarrow 1}\frac{x^3-x}{x-1}&=\lim _{x \rightarrow 1}\frac{x(x^2-1)}{x-1}\\&=\lim _{x \rightarrow 1}\frac{x(x+1)(x-1)}{x-1} \\ &=\lim _{x \rightarrow 1} x(x+1) \\ &=1(1+1)\\ &=2. \end{align*}
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