Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - Review - Exercises: 37

Answer

$0$

Work Step by Step

$\lim _{x\rightarrow 0}\dfrac {1-\sqrt {1-x^{2}}}{x}=\dfrac {\left( 1-\sqrt {1-x^{2}}\right) \left( 1+\sqrt {1-x^{2}}\right) }{x\left( 1+\sqrt {1-x^{2}}\right) }=\dfrac {x^{2}}{x\left( 1+\sqrt {1-x^{2}}\right) }=\dfrac {x}{\left( 1+\sqrt {1-x^{2}}\right) }=\dfrac {0}{1+\sqrt {1}}=0$
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