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 - Page 109: 30

Answer

$8$

Work Step by Step

$\lim _{x\rightarrow \infty }\left( \dfrac {2x+1}{x}\right) ^{3}=\left( \dfrac {\dfrac {2x}{x}+\dfrac {1}{x}}{\dfrac {x}{x}}\right) ^{3}=\left( 2+\dfrac {1}{x}\right) ^{3}=\left( 2+0\right) ^{3}=2^{3}=8$
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