$ x_{n+1}=\frac{2x_n^3}{3x_n^2-1}$ Find all $a$ such that $\{x_n\}$ has finite limit Let $\{x_n\}$ be a real sequence defined by:
$$
x_1=a \\
x_{n+1}=\frac{2x_n^3}{3x_n^2-1}
$$
For all $n=1,2,3...$ and $a$ is a real number.
Find all $a$ such that $\{x_n\}$ has finite limit when $n\to +\infty$ and find the finite limit in that cases.
I'm stuck when $a\in(-1,1)$ and $a<-1$
Please help me. Thanks

HINT: Let $$f(x)=\frac{2x^3}{3x^2-1}-x\;,$$ and show first that any limit of a sequence $\langle x_n:n\in\Bbb Z^+\rangle$ must be a zero of $f$. Then show that $f\,'(x)<0$ for all $x\in(-1,1)$ for which the function is defined. Note that $$f(-1)=f(0)=f(1)=0$$ and that $f$ has vertical asymptotes at $x=\pm\frac1{\sqrt3}$. Conclude that $f(x)<0$ for $-1<x<-\frac1{\sqrt3}$ and $0<x<\frac1{\sqrt3}$, and $f(x)>0$ for $-\frac1{\sqrt3}<x<0$ and $\frac1{\sqrt3}<x<1$. Use this information to determine the limit of $\langle x_n:n\in\Bbb Z^+\rangle$ for values of $x_1$ in $(-1,1)$.
