## Calculus 8th Edition

Published by Cengage

# Chapter 1 - Functions and Limits - Review - Exercises - Page 97: 27

#### Answer

$\dfrac{3}{2}$

#### Work Step by Step

$\lim\limits_{x \to -3}\dfrac{x^2-9}{x^2+2x-3}$ Factor: $\lim\limits_{x \to -3}\dfrac{(x+3)(x-3)}{(x+3)(x-1)}$ The $x+3$ cancels out: $\lim\limits_{x \to -3}\dfrac{x-3}{x-1}$ Plug in -3: $\lim\limits_{x \to -3}\dfrac{-3-3}{-3-1}=\dfrac{-6}{-4}=\dfrac{3}{2}$

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