Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.1 - Sequences - Exercises 10.1 - Page 570: 91



Work Step by Step

Consider $a_{n+1}=\dfrac{72}{1+a_n}$ and $a_1=2$ suppose $l=\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\dfrac{72}{1+a_n}$ Thus, $l=\dfrac{72}{1+l}$ or, $l^2+l-72=0 \implies (l+9)(l-8)=0 \implies l=-9 $ or, $8$ As we are given that $a_1=2$; and $2 \gt 0$ Hence, the limit of the sequence is $l=8$.
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