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: 46

Answer

$-\frac{1}{25}=-0.04$

Work Step by Step

$\lim _{h\rightarrow 0}\dfrac {\dfrac {1}{5+h}-\dfrac {1}{5}}{h}=\dfrac {\dfrac {5-\left( 5+h\right) }{5\times \left( 5+h\right) }}{h}=\dfrac {-h}{5\times \left( 5+h\right) \times h}=\dfrac {-1}{5\times \left( 5+h\right) }=\dfrac {-1}{5\times (5+0)}=-\frac{1}{25}$
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