University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 488: 38


$\lim\limits_{n \to \infty} a_n=6$ and {$a_n$} is convergent.

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (2-\dfrac{1}{2^n})(3+\dfrac{1}{2^n})$ $=\lim\limits_{n \to \infty} (2-\dfrac{1}{2^n}) \cdot \lim\limits_{n \to \infty} (3+\dfrac{1}{2^n})$ $=2 \cdot 3$ or, $=6$ Thus, $\lim\limits_{n \to \infty} a_n=6$ and {$a_n$} is convergent.
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