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

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

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: 38

Answer

$$4$$

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: $$ \lim\limits_{x \to -1} \dfrac{(x^3+x^2+3x+3)}{(x^4+x^3+2x+2)}=\lim\limits_{x \to -1} \dfrac{x^2(x-1)+3(x+1)}{x^3(x+1)+2(x-1)} \\=\dfrac{\lim\limits_{x \to -1} (x+1)(x^2+3)}{\lim\limits_{x \to -1} (x +1)(x^3+2)}\\=\dfrac{\lim\limits_{x \to -1} (x^2+3)}{\lim\limits_{x \to -1} (x^3+2)} \\=\dfrac{1+3}{-1+2} \\=4$$
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