Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.2 - The Limit of a Function - 2.2 Exercises - Page 93: 19


$\lim\limits_{x \to 3}\frac{x^{2}-3x}{x^{2}-9}=1/2$

Work Step by Step

For the numerator $x^{2}-3x= x(x-3)$ For the denominator $x^{2}-9= (x-3)(x+3)$ Cancel (x-3) We have $\lim\limits_{x \to 3}\frac{x}{x+3}=3/(3+3)=1/2$
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