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: 26

Answer

$-3$

Work Step by Step

We want to find $\lim\limits_{x \to 2}\dfrac{x^2-7x+10}{x-2}$, 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 $\dfrac{x^2-7x+10}{x-2}=\dfrac{(x-2)(x-5)}{x-2}=x-5$ for all $x \neq 2.$ Thus $\lim\limits_{x \to 2}\dfrac{x^2-7x+10}{x-2}=\lim\limits_{x \to 2}x-5=2-5=-3.$
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