Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.3 Techniques for Computing Limits - 2.3 Exercises - Page 78: 63

Answer

$5$

Work Step by Step

$\lim _{x\rightarrow 1}\dfrac {\sqrt {10x-9}-1}{x-1}=\lim _{x\rightarrow 1}\dfrac {\sqrt {10x-9}-1}{\left( x-1\right) }\times \dfrac {\sqrt {10x-9}+1}{\sqrt {10x-9}+1}=\dfrac {\left( \sqrt {10x-9}\right) ^{2}-1^{2}}{\left( x-1\right) \sqrt {10x-9}+1}=\dfrac {10\left( x-1\right) }{\left( x-1\right) \left( \sqrt {10x-9}+1\right) }=\dfrac {10}{\sqrt {10-9}+1}=\dfrac {10}{2}=5 $
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