Calculus: Early Transcendentals (2nd Edition)

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

Chapter 8 - Sequences and Infinite Series - Review Exercises - Page 658: 4

Answer

$e^6$

Work Step by Step

$\lim\limits_{n \to \infty} (1+\dfrac{3}{n})^{2n}=\lim\limits_{n \to \infty} [(1+\dfrac{3}{n})^{n}]^2\\=(e^3)^2 \\=e^6$
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