## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson

# Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Chapter Review - Review Exercises - Page 925: 6

#### Answer

$-\dfrac{1}{4}$

#### Work Step by Step

We will use the following limit rules for this problem: $\ Rule \ - \ 1 : \lim\limits_{x \to a} A(x)=A(a)$ $\ Rule \ - \ 2: \lim\limits_{x \to a} \dfrac{A(x)}{B(x)}=\dfrac{\lim\limits_{x \to a} A(x)}{\lim\limits_{x \to a} B(x)}$ Apply the given rules: $\lim\limits_{x\to -1}\dfrac{x^2+x+2}{x^2-9}=\dfrac{\lim\limits_{x\to -1} (x^2+x+2)}{\lim\limits_{x\to -1}(x^2-9) } \\=\dfrac{(-1)^2+(-1)+2}{(-1)^2-9}\\=-\dfrac{1}{4}$

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