Answer
$\dfrac {1}{2}$
Work Step by Step
$\lim _{x\rightarrow 1}\dfrac {1-\sqrt {x}}{1-x}=\dfrac {1-\sqrt {x}}{\left( 1-\sqrt {x}\right) \left( 1+\sqrt {x}\right) }=\dfrac {1}{1+\sqrt {x}}=\dfrac {1}{1+\sqrt {1}}=\dfrac {1}{2}$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 2 - Practice Exercises - Page 111: 11
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