Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.3 Exercises: 86

Answer

$\lim\limits_{\Delta x\to0}\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}=\dfrac{1}{2\sqrt{x}}.$

Work Step by Step

$\lim\limits_{\Delta x\to0}\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}$ $=\lim\limits_{\Delta x\to0}\dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}\times\dfrac{\sqrt{x+\Delta x}+\sqrt{x}}{\sqrt{x+\Delta x}+\sqrt{x}}$ $=\lim\limits_{\Delta x\to0}\dfrac{(\sqrt{x+\Delta x})^2-(\sqrt{x})^2}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}$ $=\lim\limits_{\Delta x\to0}\dfrac{\Delta x}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}$ $=\lim\limits_{\Delta x\to0}\dfrac{1}{\sqrt{x+\Delta x}+\sqrt{x}}$ $=\dfrac{1}{\sqrt{x+0}+\sqrt{x}}=\dfrac{1}{2\sqrt{x}}.$
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