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: 92



Work Step by Step

Since, we are given that $a_{n+1}=\dfrac{a_n+6}{a_n+2}$ and $a_1=-1$ Consider $\lim\limits_{n \to \infty}a_{n+1}=\lim\limits_{n \to \infty}\dfrac{a_n+6}{a_n+2}=l$ and $\lim\limits_{n \to \infty} a_n=l$ Thus, $l=\dfrac{l+6}{l+2}$ or, $l^2+l-3=0$ or, $l=-3 $ or, $2$ Since, we have $a_1=-1$, which is $-1 \lt 0$ As per the Principal of Mathematical Induction $a_n \geq 0$, which means that $ l \geq 0$ Hence, the limit of the sequence is $2$
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