Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 56: 29



Work Step by Step

We want to find $\lim\limits_{x \to -2}\dfrac{-2x-4}{x^3+2x^2}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches -2 is zero, and if we try to substitute $x=-2$ directly, we get zero in both the numerator and the denominator. This means we have a common factor of $x+2$ in both the numerator and the denominator. Note that $\dfrac{-2x-4}{x^3+2x^2}=\dfrac{-2(x+2)}{x^2(x+2)}=\dfrac{-2}{x^2}$ for all $x \neq -2.$ Thus $\lim\limits_{x \to -2}\dfrac{-2x-4}{x^3+2x^2}=\lim\limits_{x \to -2}\dfrac{-2}{x^2}=\dfrac{-2}{(-2)^2}=\dfrac{-2}{4}=-\dfrac{1}{2}.$
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