University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Practice Exercises: 11

Answer

$\dfrac {1}{2}$

Work Step by Step

$\lim _{x\rightarrow 1}\dfrac {1-\sqrt {x}}{1-x}=\dfrac {1-\sqrt {x}}{\left( 1-\sqrt {x}\right) \left( 1+\sqrt {x}\right) }=\dfrac {1}{1+\sqrt {x}}=\dfrac {1}{1+\sqrt {1}}=\dfrac {1}{2}$
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