# Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Section 13.2 Algebra Techniques for Finding Limits - 13.2 Assess Your Understanding - Page 903: 31

$$\dfrac{7}{6}$$

#### Work Step by Step

In order to simplify the given expression, we will use the following rules. $(a) \lim\limits_{x \to a} \dfrac{p(x)}{q(x)}=\dfrac{\lim\limits_{x \to a} p(x)}{\lim\limits_{x \to a} q(x)} \\ (b) \lim\limits_{x \to a} k(x)=k(a)$ ; where $a$ is a constant. --- Thus, we have: $$\lim\limits_{x \to -3} \dfrac{x^2-x-12}{x^2-9}=\lim\limits_{x \to -3} \dfrac{(x+3)(x-4)}{(x+3)(x-3)} \\=\dfrac{\lim\limits_{x \to -3} x-4}{\lim\limits_{x \to -3} x-3} \\=\dfrac{-3-4}{-3-3}\\=\dfrac{7}{6}$$

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