Calculus: Early Transcendentals (2nd Edition)

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

Chapter 2 - Limits - Review Exercises - Page 124: 31


$\dfrac {1}{2}$

Work Step by Step

$\lim _{x\rightarrow \infty }\dfrac {2x-3}{4x+10}=\lim _{x\rightarrow \infty }\dfrac {2-\dfrac {3}{x}}{4+\dfrac {10}{x}}=\dfrac {2-0}{4+0}=\dfrac {2}{4}=\dfrac {1}{2}$
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