Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.3 Techniques for Computing Limits - 2.3 Exercises - Page 77: 32


$\dfrac {3}{8}=0.375$

Work Step by Step

$\lim _{h\rightarrow 0}\dfrac {3}{\sqrt {16+3h}+4}=\dfrac {3}{\sqrt {16+3\times 0}+4}=\dfrac {3}{\sqrt {16}+4}=\dfrac {3}{8}$
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