Can a sequence defined by $$x_{n+1} = f(x_n)$$ exist such that it converges to $x$ but $$f(x)\ne x,$$ that is, $x$ is not a fixed point of $f$?
Motivated by this question. I wrote an answer there but something seems fishy and I cannot pinpoint what.
I see that it is obvious if you define the function pathologically (say $f(x) = (x+2/x)/2$ for real $x$ but $f(x)=0$ for imaginary x), so feel free to assume any "nice" properties for the function.
Sure. Take $f(x) = \frac{x}{2}$ for $x \neq 0$ and $f(x) =1$ for $x = 0$.
If $f$ is continuous at $x$, then $$ x=\lim_{n\rightarrow\infty}x_n=\lim_{n\rightarrow\infty}f(x_n) $$ by the definition of $x_n$. Since $f$ is continuous, we can interchange the limit and the function so that $$ \lim_{n\rightarrow\infty}f(x_n)=f\left(\lim_{n\rightarrow\infty}x_n\right)=f(x). $$ Therefore, you're looking for a function which is not continuous at $x$ in order to have the property that you're looking for.
Note that you can construct examples where the desired property holds, but $f$ is not continuous, so the precise requirements are hard to state.
An interesting related question might be if $\lim_{n\rightarrow\infty}x_n=x$ such that $x_{n+1}=f(x_n)$, can we find a continuous function $g$ that agrees with $f$ on $\{x_n\}$.