Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - Exercises - Page 167: 6



Work Step by Step

$$A=\lim\limits_{x\to1^+}\frac{x^2-9}{x^2+2x-3}$$ - Consider the numerator: $x^2-9=(x-3)(x+3)$ - Consider the denominator: $x^2+2x-3=(x-1)(x+3)$ Therefore, $$A=\lim\limits_{x\to1^+}\frac{(x-3)(x+3)}{(x-1)(x+3)}$$$$A=\lim\limits_{x\to1^+}\frac{x-3}{x-1}$$ As $x\to1^+$, $\frac{x-3}{x-1}$ approaches $\frac{1-3}{1-1}=\frac{-2}{0}=-\infty$ So, $$\lim\limits_{x\to1^+}\frac{x^2-9}{x^2+2x-3}=-\infty$$
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