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: 23



Work Step by Step

We want to find $\lim\limits_{x \to 5}\dfrac{x-5}{x^2-25}$, 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 into our function, 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 $\dfrac{x-5}{x^2-25}=\dfrac{x-5}{(x-5)(x+5)}=\dfrac{1}{x+5}$ for all $x \neq 5.$ Thus $\lim\limits_{x \to 5}\dfrac{x-5}{x^2-25}=\lim\limits_{x \to 5}\dfrac{1}{x+5}=\dfrac{1}{5+5}=\dfrac{1}{10}.$
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