Calculus (3rd Edition)

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

Chapter 2 - Limits - 2.2 Limits: A Numerical and Graphical Approach - Exercises - Page 54: 20



Work Step by Step

We have $$ \lim _{x\:\rightarrow \:3}\:\frac{x^3-2x^2-9}{x^2-2x-3}= \lim _{x\:\rightarrow \:3}\:\frac{(x-3)(x^2+x+3)}{(x-3)(x+1)}= \lim _{x\:\rightarrow \:3} \frac{x^2+x+3}{x+1}=\frac{15}{4}. $$ Where we factored the numerator and denominator, canceled out common terms and plugged in "3" for $x$.
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