We have that:
$\bullet$ $f'$ is continuous on $\mathbb{R}$,
$\bullet$$f(a)=a$,
$\bullet$ $|f'(a)<1|$
We define a sequence given $x_0 \in \mathbb{R}$ by $x_{n+1}=f(x_n)$.
We want to show that if $x_0$ is sufficiently close to a, then $\lim_{x \to \infty} x_n$ = a.
I have been working on this problem for a while, and keep dead ending. I know that $x_0$ sufficiently close to a means $x_0 \in (a-\delta, a+\delta)$ for some $\delta$. I cannot see how all of this comes together in the proof and would appreciate a little insight.
Also, I know that if I want to show a sequence converges, to a in this case, I need to show that for n>N, $|x_n-a|<\epsilon$ and can't see how to go about doing this. I don't know if a proof by induction would be helpful here or not.
Hint: Note that since $|f'(a)|<1$ and $f'$ is continuous, then $|f'(x)|<1$ for $x$ sufficiently close to $a$. Try to work with that degree of proximity.
The insight here is that you are trying to show something like the Banach fixed point theorem and we want to take advantage of the conditions to locally see $f$ as a contraction.