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: 41

Answer

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

Work Step by Step

$\lim\limits_{x\to4^-}\dfrac{\sqrt{x}-2}{x-4}=\lim\limits_{x\to4^-}[\dfrac{\sqrt{x}-2}{x-4}\times\dfrac{\sqrt{x}+2}{\sqrt{x}+2}]$ $=\lim\limits_{x\to4^-}\dfrac{(\sqrt{x})^2-(2)^2}{(x-4)(\sqrt{x}+2)}=\lim\limits_{x\to4^-}\dfrac{(x-4)}{(x-4)(\sqrt{x}+2)}$ $=\lim\limits_{x\to4^-}\dfrac{1}{\sqrt{x}+2}=\dfrac{1}{\sqrt{4^-}+2}=\dfrac{1}{4}.$
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