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: 51

Answer

$\dfrac {1}{8}=0.125 $

Work Step by Step

$\lim _{h\rightarrow 0}\dfrac {\sqrt {16+h}-4}{h}=\dfrac {\sqrt {16+h}-4}{h}\times \dfrac {\sqrt {16+h}+4}{\sqrt {16+h}+4}=\dfrac {\left( \sqrt {16+h}\right) ^{2}-4^{2}}{h\times \left( \sqrt {16+h}+4\right) }=\dfrac {16+h-16}{h\times \left( \sqrt {16+h}+4\right) }=\dfrac {1}{\sqrt {16+h}+4}=\dfrac {1}{\sqrt {16}+4}=\dfrac {1}{8} $
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