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: 25



Work Step by Step

We want to find $\lim\limits_{x \to -5}\dfrac{x^2+3x-10}{x+5}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches -5 is zero, and if we try to substitute $x=-5$ directly, we get zero in both the numerator and the denominator. This means we have a common factor of $x+5$ in both the numerator and the denominator. Note that $\dfrac{x^2+3x-10}{x+5}=\dfrac{(x+5)(x-2)}{x+5}=x-2$ for all $x \neq -5.$ Thus $\lim\limits_{x \to -5}\dfrac{x^2+3x-10}{x+5}=\lim\limits_{x \to -5}x-2=-5-2=-7.$
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