Calculus: Early Transcendentals (2nd Edition)

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

Chapter 2 - Limits - 2.6 Continuity - 2.6 Exercises: 34

Answer

$\dfrac {1}{2} =0.5 $

Work Step by Step

$\lim _{x\rightarrow 0}\left( \dfrac {x}{\sqrt {16x+1}-1}\right) ^{\dfrac {1}{3}}=\left( \dfrac {x\left( \sqrt {16x+1}+1\right) }{\left( \sqrt {16x+1}-1\right) \left( \sqrt {16x+1}-1\right) }\right) ^{\dfrac {1}{3}}=\left( \dfrac {x\left( \sqrt {16x+1}+1\right) }{16x+1-1}\right) ^{\dfrac {1}{3}} =\left( \dfrac {x\left( \sqrt {16x+1}+1\right) }{16x}\right) ^{\dfrac {1}{3}}={(\dfrac {\sqrt {16x+1}+1}{16})}^\frac{1}{3}={(\dfrac {\sqrt {16\times 0+1}+1}{16})}^\frac{1}{3}=\left( \dfrac {2}{16}\right) ^{\dfrac {1}{3}}=\left( \dfrac {1}{8}\right) ^{\dfrac {1}{3}}=\dfrac {1}{2} =0.5 $
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