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 59: 86

Answer

$-4$

Work Step by Step

To find the $\lim\limits_{x \to -1}\frac{x^3 - x^2 - 5x - 3}{(x + 1)^2}$ when we apply limit directly we get the indeterminate form. So we calculate the given limit as follows First, factorize the numerator: \[ x^3 - x^2 - 5x - 3 = (x + 1)(x^2 - 2x - 3) = (x + 1)(x - 3)(x + 1) \] Now, the expression becomes: \[ \lim_{x \to -1} \frac{(x + 1)(x - 3)(x + 1)}{(x + 1)^2} \] Cancel out the common factor \( (x + 1) \) from the numerator and denominator: \[ \lim_{x \to -1} \frac{(x - 3)(x + 1)}{x + 1} \] Now, simplify further: \[ \lim_{x \to -1} (x - 3) \] Now, substitute \( x = -1 \): \[ (-1 - 3) = -4 \] So, $\lim\limits_{x \to -1}\frac{x^3 - x^2 - 5x - 3}{(x + 1)^2}=-4$.
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