TITLE: How prove this limits is exsit $\displaystyle\lim_{n\to\infty}x_{n}$
QUESTION [5 upvotes]: let $f:[a,b]\to [a,b]$ be Continuous function,Assmue that sequence $\{x_{n}\}(n\ge 0)$ such
  $$x_{0}=x,x_{1}=f(x_{0}),x_{2}=f(x_{1}),\cdots,x_{n+1}=f(x_{n}),\forall n\in N^{+}$$
  and $$\lim_{n\to\infty}(x_{n+1}-x_{n})=0$$
  show that:
  $$\lim_{n\to\infty}x_{n}$$ is exsit

My idea: first  we note
$$x_{n+1}-x_{n}=f(x_{n})-x_{n}$$
so
we have
$$\lim_{n\to\infty}(f(x_{n})-x_{n})=0$$
then I can't prove this limits is exsit?I fell can't easy prove it.?

REPLY [2 votes]: WLOG $f:[0,1]\to[0,1]$. Let $a=\underline\lim x_n$, $b=\overline\lim x_n$. Assume $a<b$, clearly $f$ can't be identically $x$ on any subinterval of $[a,b]$, as the sequence would stop due to the limit condition. So there's some $c\in (a,b)$ so that WLOG $f(c)>c$. 
$f$ is continuous in $c$, so let $\varepsilon>0$ be so small that $f(x)>c+\varepsilon$ for all $x\in(c-\varepsilon,c+\varepsilon)$ and $c-\varepsilon>a$, $c+\varepsilon<b$. Together with the limit condition this guarantees that $x_n$ can never cross the $(c-\varepsilon,c+\varepsilon)$ interval on a path from $b$ to $a$ for sufficiently large $n$, which is a contradiction.