University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises - Page 66: 22


$\lim\limits_{h \to 0}\frac{\sqrt {5h+4} -2}{h} = 1.25=\frac{5}{4}$

Work Step by Step

$\frac{\sqrt (5h+4) -2}{h}= \frac{(\sqrt (5h+4) -2)(\sqrt (5h+4) +2)}{h(\sqrt (5h+4) +2)}$ = $\frac{ (5h+4) -4)}{h(\sqrt (5h+4) +2)}$ = $\frac{ (5h)}{h(\sqrt (5h+4) +2)}$ = $\frac{5}{(\sqrt (5h+4) +2)}$ Now, we apply the property of limits: $\lim\limits_{h \to 0}\frac{5}{(\sqrt (5h+4) +2)} = \frac{\lim\limits_{h \to 0}5}{\lim\limits_{h \to 0}(\sqrt (5h+4) +2)}$ = $\frac{5}{4}$ = 1.25
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