Let us compute:
$$
x-f(x)=x-\frac{x}{1+bx}=\frac{bx^2}{1+bx}.
$$
So we see that $x-f(x)>0$ for all $x>0$.
An easy induction proof shows that $x_n>0$ for all $n$.
So for all $n$,
$$
x_n-f(x_n)=x_n-x_{n+1}>0.
$$
Hencee, indeed, the sequence $(x_n)$ is decreasing and bounded below by $0$.
So it converges to some limit $L\geq 0$.
Now since $f$ is continuous, we must have at the limit
$$
L=f(L)\quad\Leftrightarrow\quad L=\frac{L}{1+bL}\quad\Leftrightarrow\quad L=0.
$$
So $(x_n)$ converges to $0$.
assume that it convergences to x. In which case, $x = \frac{1}{4-x}$
solve for x. There are two possible solutions... one of those will prove to be unstable.
Now that you know what x should equal it should be simple to show that $x_1>x_2>...>x$ your sequence is bounded below and monotonically decreasing and therefore convergent.
Suppose {$x_n$} converges to $x$.
$x = \frac{1}{4-x}\\
4x - x^2 = 1\\
- (x^2 -4x + 1) =0\\
- (x - 2+\sqrt3)(x - 2-\sqrt3)=0\\
$
I claim that
$\forall x_n\in(2-\sqrt3,2+\sqrt3), 2-\sqrt3<x_{n+1}<x_n$
to show that $x_{n+1}<x_n$, we can show that $x_n - x_{n+1}>0$
$x_n - x_{n+1} = x_n - \frac{1}{4-x_n} = \frac{-x_n^2 + 4x_n - 1}{4-x_n} $
and $\frac{-x_n^2 + 4x_n - 1}{4-x_n} > 0$ when x is in the interval.
And to show that $x_{n+1} > 2-\sqrt3$, when $x_n>2-\sqrt3, 4-x_n < 2+\sqrt3$ and $\frac{1}{4-x_n} > \frac{1}{2+\sqrt3}.$
$x_2 \in(2-\sqrt3,2+\sqrt3)$ then for all $n>2, 2-\sqrt3<x_{n+1}<x_n$
$x_n$ is monotonically decreasing and is bounded below.
Best Answer
If the proposed sequence converges, then it is bounded.
Having said that, let us take the limit:
\begin{align*} \lim_{n\to\infty}x_{n} & = \lim_{n\to\infty}\frac{5n^{6} + 6}{(n^{4} + 1)(n^{2} - 2)}\\\\ & = \lim_{n\to\infty}\frac{5 + 6/n^{6}}{(1 + 1/n^{4})(1 - 2/n^{2})}\\\\ & = \frac{5 + 0}{(1 + 0)(1 - 0)} = 5 \end{align*}
Since it is convergent, then it is bounded.
Hopefully this helps!