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: 24


$$-b $$

Work Step by Step

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