## Calculus: Early Transcendentals 8th Edition

a) $\lim\limits_{n \to \infty}a_{n} =\lim\limits_{n \to \infty}a_{n+1}$ b) $L=\frac{\sqrt 5 -1}{2}$
a) Convergent sequence {$a_{n}$}. If a sequence is convergent then it has a limit. Suppose the limit of the sequence {$a_{n}$} is $L$. Thus $\lim\limits_{n \to \infty} = L$ or $a_{n} → L$ as $n → \infty$ That is, if, for every $ε \gt 0$ ther is a corresponding integer N such that If $n \gt N$, then $|a_{n} - L| \lt ε$ $n \gt N$ so, $n+1 \gt N$, So, $|a_{n+1}-L| \lt ε$ Hence $\lim\limits_{n \to \infty}a_{n+1} = L$ Therefore $\lim\limits_{n \to \infty}a_{n} =\lim\limits_{n \to \infty}a_{n+1}$ b) From part (a) $\lim\limits_{n \to \infty}a_{n} =\lim\limits_{n \to \infty}a_{n+1}$ $\lim\limits_{n \to \infty}a_{n} = \lim\limits_{n \to \infty}(\frac{1}{1+a_{n}})$ Use $a_{n}=\frac{1}{1+a_{n}}$ $\lim\limits_{n \to \infty}a_{n} = \frac{1}{\lim\limits_{n \to \infty}(1+a_{n})}$ $\lim\limits_{n \to \infty}a_{n} = \frac{1}{1+\lim\limits_{n \to \infty}a_{n}}$ $L=\frac{1}{1+L}$ use $\lim\limits_{n \to \infty}a_{n}=L$ $L^{2}+L=1$ $L^{2}+L-1=0$ $L=\frac{-1±\sqrt (1^{2}-4(1)(-1))}{2(1)}$ $L=\frac{-1±\sqrt (1+4)}{2}$ $L=\frac{-1±\sqrt (5)}{2}$ $L=\frac{-1+\sqrt (5)}{2}$ or $L=\frac{-1-\sqrt (5)}{2}$ The sequence can be defined as $a_{1}=1$ and $a_{n}=\frac{1}{(1+a_{n})}$ for $n \geq 1$ The sequence is $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, ...$ and all the terms are positive. As $n→ \infty$, the limit $L$ will be a very small positive number but cannot be negative. Hence, the limit of the sequence {$a_{n}$} is $L=\frac{\sqrt 5 -1}{2}$