Prove that if F(x) is continuous and $F^n(x_0)$ converges to $y_1$ for n even and $y_2$ for n odd then {$y_1$,$y_2$} is a 2-cycle for F(x).
note: $F^n(x_0)$ means applying the function $F(x)$ $n$ times. (I used $n=2k$ for $n$ even and $2k+1$ for $n$ odd)
Attempt: I read that for a 2-cycle we must solve $F^2(x)=x$ but how does that help in the general case?
$y_1=lim_nF^{2n}(x_0)$, since $F$ is continuous, $F(y_1)=lim_nF(F^{2n}(x_0))=lim_nF^{2n+1}(x_0)=y_2$, similarly $y_2=lim_nF^{2n+1}(x_0)$, since $F$ is continuous, $F(y_2)=lim_nF(F^{2n+1}(x_0)=lim_nF^{2(n+1)}(x_0)=y_1$. $F(y_1)=y_2, F(y_2)=y_1$.