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: 96



Work Step by Step

Since, we are given that $a_{n+1}=12-\sqrt {a_n}$ Consider $\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}12-\sqrt {a_n}=l$ and $\lim\limits_{n \to \infty} a_n=l$ Thus, $l=12-\sqrt l$ or, $l^2-25l+144=0$ or, $l=9 $ or, $16$ Since, we have $12-\sqrt {a_n} \lt 12$ Hence, the limit of the sequence is $9$
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