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: 56

Answer

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

Work Step by Step

$f(x)=\dfrac{\sqrt{2+x}-\sqrt{2}}{x}=\dfrac{\sqrt{2+x}-\sqrt{2}}{x}\times\dfrac{\sqrt{2+x}+\sqrt{2}}{\sqrt{2+x}+\sqrt{2}}$ $=\dfrac{(\sqrt{2+x})^2-(\sqrt{2})^2}{x(\sqrt{2+x}+\sqrt{2})}=\dfrac{x}{x(\sqrt{2+x}+\sqrt{2})}=\dfrac{1}{\sqrt{2+x}+\sqrt{2}}=g(x).$ The function $g(x)$ agrees with the function $f(x)$ at all points except $x=0$. Therefore we find the limit as x approaches $0$ of $f(x)$ by substituting the value into $g(x)$. $\lim\limits_{x\to0}\dfrac{\sqrt{2+x}-\sqrt{2}}{x}=\lim\limits_{x\to0}\dfrac{1}{\sqrt{2+x}+\sqrt{2}}=\dfrac{1}{\sqrt{0+2}+\sqrt{2}}=\dfrac{1}{2\sqrt{2}}.$
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