Examining sequence convergence with parameter $a$? For $a\in R$, let $x_1=a$ and $x_{n+1}$=$-\frac{6x_n^2+4x_n}{x_n^2-2x_n+4}$. Examine the convergence of the sequence ${(x_n)}_{n=1}^{\infty}$ for different values of $a$. Also find $\lim_{n\to\infty}x_n$ whenever it exist.
I know that if the sequence converges to $x$ then it must satisfy $x$=$-\frac{6x^2+4x}{x^2-2x+4}$, that is $x=0$. 
$x_{n+1}$$-$$x_n$=-$\frac{x_n(x_n^2+4x_n+8)}{x_n^2-2x_n+4}$ and should use it to find out whether the sequence is monotone . But i don't know how to do it and how to use $a$ in all this and find $a$ so that the sequence converges. Can someone explain it ? Thanks. 
I can't find any answer. I'm stuck. If someone can help ,please do it:)

The only possible real limit is
$0$,
since all the other roots of
$x
=-\frac{6x^2+4x}{x^2-2x+4}
$
are complex.
Around zero,
it can't be monotone
since
$x_{n+1}
=-\frac{x_n(6x_n+4)}{x_n^2-2x_n+4}
$
has the opposite sign
of $x_n$.
What is needed,
imho,
is to write
$x_{n+2} = \text{some function}(x_n)$.
This would probably show convergence,
but I'm too lazy right now
to do it.
