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 103: 51

Answer

$c=7$

Work Step by Step

$\lim\limits_{t \to 2^-}B(t)=\lim\limits_{t \to 2^-}(4-\frac{1}{2}t)$ $\lim\limits_{t \to 2^+}B(t)=\lim\limits_{t \to 2^+}\sqrt{t+c}$ For $\lim\limits_{t \to 2}B(t)$ to exist, $\lim\limits_{t \to 2^-}B(t)=\lim\limits_{t \to 2^+}B(t)$ $\lim\limits_{t \to 2^-}(4-\frac{1}{2}t)=\lim\limits_{t \to 2^+}\sqrt{t+c}$ $4-\frac{2}{2}=\sqrt{2+c}$ $2+c=3^2$ $c=7$
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