Let $(x_n)$ be a sequence of real numbers such that $x_{n+1}=f(x_n)$ for all natural numbers $n$, where $f$ is a function from $\mathbb{R}$ to $\mathbb{R}$. And suppose that $(x_n)$ converges to some real number $x$. Then my question is, is it necessarily true that $f(x)=x$?
Clearly it's true if $f$ is continuous, since $f(x_n)\rightarrow f(x)$ and $f(x_n)=x_{n+1}\rightarrow x$. But what about the general case?
Best Answer
$f$ is merely "a function"? Sure. Take your favorite case where you have a continuous function $f$, and convergence to a fixed point $c$, then change the value $f$ at that one point $c$, so that it is no longer a fixed point..
Example... $f(x) = x/2$ for $x \ne 0$ and $f(0) = \pi$. Then $x_n = 1/2^n$ satisfies $x_{n+1} = f(x_n)$ and $x_n \to 0$, but $0$ is not a fixed point of $f$.