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: 94



Work Step by Step

Since, we are given that $a_{n+1}=\sqrt {8+2a_n}$ and $a_1=0$ Consider $\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\sqrt {8+2a_n}=l$ and $\lim\limits_{n \to \infty} a_n=l$ Thus, $l=\sqrt {8+2l}$ or, $l^2-2l-8=0$ or, $l=-2 $ or, $4$ Since, we have $a_1=0$, which is $0 \gt 0$ and $a_2=\sqrt 8 \gt 0$ As per the Principal of Mathematical Induction $a_n \geq 0$, which means that $ l \geq 0$ Hence, the limit of the sequence is $4$
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