Calculus: Early Transcendentals (2nd Edition)

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

Chapter 8 - Sequences and Infinite Series - 8.2 Sequences - 8.2 Exercises: 12

Answer

$\lim\limits_{n \to \infty}\frac{2e^n + 1}{e^n} = 2$

Work Step by Step

$\lim\limits_{n \to \infty}\frac{2e^n + 1}{e^n} = \lim\limits_{n \to \infty}\frac{\frac{2e^n}{e^n}+ \frac{1}{e^n}}{\frac{e^n}{e^n}} = \lim\limits_{n \to \infty}\frac{2+ \frac{1}{e^n}}{1} = \lim\limits_{n \to \infty}\frac{2+0}{1} = 2$
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