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: 30


$$ 3-a$$

Work Step by Step

By substituting, we get the intermediate form $\frac{0}{0}$. So by factorizing the numerator, we have $$ \lim _{x \rightarrow 1} \frac{ x^3-ax^2+ax-1} {x-1 }=\lim _{x \rightarrow 1} \frac{( x^3- 1)-ax(x-1)} {x-1 }\\ =\lim _{x \rightarrow 1} \frac{( x- 1)(x^2+x+1)-ax(x-1) }{x-1 }=\lim _{x \rightarrow 1} x^2+x -ax+1\\ =1+1-a+1=3-a. $$
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