which user answer is correct ?. It is given that the function $f(x) = \frac{x^3 + 1}{3}$ has three fixed points $\alpha$,$\beta$,$\gamma$ in $(-2,-1), (0, 1)$ and $(1, 2)$ respectively. A sequence of real numbers is defined as $$x_1 =\gamma - 0.01\qquad , \qquad x_{n+1} = f(x_n) \enspace\forall n = 1, 2, 3,\cdots.$$ Given that the sequence converges, find 
$\displaystyle\lim_{n\rightarrow\infty}x_n$.
My attempt :  I have  found the  two answer , Im confused that  which one  correct ?
first anwser by jose carlos sir 
$2$nd answer  by Muhammad sir

Link to first: find $\lim_{n→∞}x_n$?..→∞x-n
Link to second: Convergence of sequence involving fixed points of a function
Here, I'm confused  that  which  answer is correct ?
 A: We can use a discrete version of the derivative, namely $x_{n+1}-x_n$, to talk about the dynamics of this system. The fixed points are equilibria of the system, that is to say that their discrete "derivative" vanishes. So once you are at a fixed point, your value doesn't change anymore. 
Since $\gamma-0.01\in(\beta,\gamma)$ let's take a look at the values of the "derivative" on that interval (the "derivative" is only $0$ at the fixed points, so intermediate value theorem tells us it must have a definite sign between them). The only number we know for sure we can test that is between the two points is $1$. Plugging in,
$$\frac{1^3+1}{3}-1 = -\frac{1}{3} < 0$$
so the sequence is always decreasing for values on that interval. But do the values on the interval stay within that interval? How do we know that it will continue sliding down to $\beta$ instead of, say, hopping over to $-1$?
We can note that $\frac{x^3+1}{3}$ is a monotonically increasing function, which means that $a\leq b \implies f(a)\leq f(b)$. Using this we can deduce that for any $x_n\in(\beta,\gamma)$:
$$\beta \leq x_n \leq \gamma \implies f(\beta) \leq f(x_n) \leq f(\gamma) \implies \beta \leq x_{n+1} \leq \gamma$$
because $\beta$ and $\gamma$ were fixed points of $f$. Therefore we have proved that the sequence is always decreasing and always confined to the interval $(\beta,\gamma)$, so the limit has to be $\beta$. 
(One can prove by contradiction that the sequence cannot stop "before" it reaches $\beta$, since the sequence must still decrease for $\beta+\epsilon$).
A: You can answer experimentally. By solving the cubic, $\gamma-0.01\approx 1.522088886238$ and the next iterates are
$$1.4781786918566,\\1.4099461806973,\\1.2676333392617,\\1.0123175885187,\\0.6791366011261,\\0.4377452710573,\\0.3490536023807,\\0.347509379493,\\0.3473220649546,\\\cdots$$
We have $\beta\approx0.3473220649546$.
This is not a formal proof as finite arithmetic might trick us, but the chances are very very low.
