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 - Review Exercises: 21

Answer

$\lim\limits_{x\to4}\dfrac{\sqrt{x-3}-1}{x-4}=\dfrac{1}{2}.$

Work Step by Step

$f(x)=\dfrac{\sqrt{x-3}-1}{x-4}=\dfrac{\sqrt{x-3}-1}{x-4}\times\dfrac{\sqrt{x-3}+1}{\sqrt{x-3}+1}$ $=\dfrac{(\sqrt{x-3})^2-(1)^2}{(x-4)(\sqrt{x-3}+1)}=\dfrac{(x-4)}{(x-4)(\sqrt{x-3}+1)}$ $=\dfrac{1}{\sqrt{x-3}+1}=g(x)$ The function $g(x)$ agrees with the function $f(x)$ at all points except $x=4$. Therefore we find the limit as x approaches $4$ of $f(x)$ by substituting the value into $g(x)$. $\lim\limits_{x\to4}\dfrac{\sqrt{x-3}-1}{x-4}=\lim\limits_{x\to4}\dfrac{1}{\sqrt{x-3}+1}=\dfrac{1}{\sqrt{4-3}+1}=\dfrac{1}{2}.$
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