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 58: 77

Answer

$7$

Work Step by Step

Given $\lim\limits_{x \to 4}\frac{f(x)-5}{x-2}=1$, we have $\frac{\lim\limits_{x \to 4}f(x)-5}{4-2}=1$ thus $\lim\limits_{x \to 4}f(x)=(4-2)+5=7$
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