Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.7 Exercises - Page 564: 36

Answer

$\infty$

Work Step by Step

$\lim _{x\rightarrow \infty }\dfrac {e^{x/2}}{x}=\dfrac {\dfrac {d}{dx}\left( e^{x/2}\right) }{\dfrac {d}{dx}\left( x\right) }=\dfrac {\dfrac {1}{2}e^{x/2}}{1}=\dfrac {1}{2}e^{x /2}=\infty $
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