I was asked the following question:
let $g(x)=\frac{30}{1+x}$. Notice that $g(5)=5$.
Is there an $\epsilon >0$ such that the series $\{x_k\}_{k=0}^{\infty}$ defined by $x_{k+1}=g(x_k)$ and $\forall j: x_j \in [5-\epsilon,5+\epsilon]$ converges to $5$?
I'm not entirely sure what to do, but however, I do know we have this theorem:
Theorem: let $g: [a,b] \to [a,b]$ be continuous. Then there is $c \in [a,b]$ such that $g(c)=c$. Furthermore, if $\forall x \in [a,b]: |g'(x)|<1$ then $c$ is the only such value that $g(c)=c$, and the iterative method $x_{n+1}=g(x_n)$ converges to $c$.
Problem is, I showed that an epsilon such that $\forall x \in [5-\epsilon,5+\epsilon]: g(x) \in [5-\epsilon,5+\epsilon]$ does not exist, and so this theorem does not help me.
I think the answer is that there is no such epsilon that the series converges, but this is merely speculation.
Try to solve the inequality $-1<g'(x)<1$; I think you should find, for example, that the inequality is true whenver $x>-1+\sqrt{30}$. Thus $\varepsilon = 5-(-1+\sqrt{30})\approx 0.522$ should do it for you.