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: 54

Answer

$\lim\limits_{x\to3}\dfrac{\sqrt{x+1}-2}{x-3}=\dfrac{1}{4}.$

Work Step by Step

$f(x)=\dfrac{\sqrt{x+1}-2}{x-3}=\dfrac{\sqrt{x+1}-2}{x-3}\times\dfrac{\sqrt{x+1}+2}{\sqrt{x+1}+2}$ $=\dfrac{(\sqrt{x+1})^2-(2)^2}{(x-3)(\sqrt{x+1}+2)}=\dfrac{(x-3)}{(x-3)(\sqrt{x+1}+2)}$ $=\dfrac{1}{\sqrt{x+1}+2}=g(x).$ The function $g(x)$ agrees with the function $f(x)$ at all points except $x=3$. Therefore, we find the limit as x approaches $3$ of $f(x)$ by substituting the value into $g(x)$. $\lim\limits_{x\to3}\dfrac{\sqrt{x+1}-2}{x-3}=\lim\limits_{x\to3}\dfrac{1}{\sqrt{x+1}+2}=\dfrac{1}{\sqrt{3+1}+2}=\dfrac{1}{4}.$
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