University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises - Page 69: 86

Answer

We can guess here that $$\lim_{x\to-1}\frac{x^3-x^2-5x-3}{(x+1)^2}=-4$$

Work Step by Step

$$\lim_{x\to-1}\frac{x^3-x^2-5x-3}{(x+1)^2}$$ (a) The graph is shown below. (b) Looking at the graph, as $x\to-1$, we see that $\frac{x^3-x^2-5x-3}{(x+1)^2}$ approaches as close as $-4$. Therefore, we can guess here that $$\lim_{x\to-1}\frac{x^3-x^2-5x-3}{(x+1)^2}=-4$$
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