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: 54

Answer

Converges to $e^{-1}$

Work Step by Step

Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (1-\dfrac{1}{n})^{n}$ Since, $\lim\limits_{n \to \infty} (1+\dfrac{x}{n})^{n}=e^x$ when $x \gt 0$ So, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (1-\dfrac{1}{n})^{n}=\lim\limits_{n \to \infty} (1+(-\dfrac{1}{n}))^{n}=e^{-1}$ Hence, $\lim\limits_{n \to \infty} a_n=e^{-1}$ and {$a_n$} is convergent and converges to $e^{-1}$
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