Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.3 - Calculating Limits Using the Limit Laws - 2.3 Exercises - Page 104: 59

Answer

$\lim\limits_{x \to 1}f(x)=8$

Work Step by Step

$\lim\limits_{x \to 1}\frac{f(x)-8}{x-1}=10$ $\frac{\lim\limits_{x \to 1} (f(x)-8)} {\lim\limits_{x \to 1} (x-1)}=10$ $\lim\limits_{x \to 1} (f(x)-8)=10{\lim\limits_{x \to 1} (x-1)}=10\times (1-1) =0$ $\lim\limits_{x \to 1} f(x)-\lim\limits_{x \to 1} 8=0$ $\therefore \lim\limits_{x \to 1} f(x)=\lim\limits_{x \to 1} 8=8$
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