Calculus: Early Transcendentals (2nd Edition)

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

Chapter 2 - Limits - Review Exercises: 10

Answer

$\lim\limits_{h \to 0}{\frac{\sqrt {5x+5h}-\sqrt {5x}}{h}} =\frac{5\sqrt {5x}}{10x}$

Work Step by Step

$\lim\limits_{h \to 0}{\frac{\sqrt {5x+5h}-\sqrt {5x}}{h}} =\lim\limits_{h \to 0}{\frac{\sqrt {5x+5h}-\sqrt {5x}}{h}\times\frac{\sqrt {5x+5h}+\sqrt {5x}}{\sqrt {5x+5h}+\sqrt {5x}}}=\lim\limits_{h \to 0}{\frac{5x+5h-5x}{h}} \times \frac{1}{\sqrt {5x+5h}+\sqrt {5x}}=\lim\limits_{h \to 0}{5\times\frac{1}{\sqrt {5x+5h}+\sqrt {5x}}}=\lim\limits_{h\to 0}{\frac{5}{\sqrt {5x+5h}+\sqrt {5x}}} =\frac{5}{\lim\limits_{h\to 0}{ \sqrt {5x+5h}+\sqrt {5x}}}=\frac{5}{{ \sqrt {5x+\lim\limits_{h\to 0}{5h}}+\sqrt {5x}}}=\frac{5}{2\sqrt {5x}}=\frac{5}{2\sqrt {5x}}\times\frac{\sqrt {5x}}{\sqrt {5x}}=\frac{5\sqrt {5x}}{10x}$
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