Let $f(x)=(3x^2-2x^3)^{1/3}$ and $x_0 \in (0,1)$, $x_{n+1}=f(x_n)$. Prove sequence converges. Let $f:\mathbb R \rightarrow \mathbb R $, $f(x)=(3x^2-2x^3)^{1/3}$.
Prove that the sequence $(x_n)$, defined by $x_0 \in (0,1)$ and $x_{n+1}=f(x_n)$, converges and calculate $\lim_{n \to \infty}x_n$.  
I need some help with this exercise. I studied the monotonicity of $f(x)$ and it is increasing on $(0,1)$. Next, I wrote the inequality that we are given upfront:
$$0 \lt x_0 \lt 1$$
Next, I applied function $f$ to the inequality, but (I guess?) the bounds do not modify at all since $f(0)=0$ and $f(1)=1$ so we end up with:
$$0 \lt x_n \lt 1$$
I do not know how to find the requested limit, that I hope you can help me with, and also I need to clarify this: The sequence $x_n$ is bounded by $0$ and $1$ and also increasing. Is that enough to prove its convergence? 
The second question: Are the above limit and the number the sequence converges to the same?
 A: First we show that for $x\in (0,1)$ $$1\ge (3x^2-2x^3)^{1\over 3}\ge x^{2\over 3}\ge 0$$by showing that $$3x^2-2x^3\ge x^2$$in this interval which is obvious since for any $x\in (0,1)$ we have $x>x^2>x^3>\cdots $
By substitution we obtain $$0\le (x_n)^{2\over 3}\le x_{n+1}=\sqrt[3]{3x_n^2-2x_n^3}<1$$applying this relation recursively $n$ times we have $$(x_0)^{\left({2\over 3}\right)^n}\le x_{n+1}<1$$ since $$\lim_{n\to \infty}(x_0)^{\left({2\over 3}\right)^n}=1$$therefore $$\lim_{n\to \infty} x_{n+1}=1$$ and the result is proved.
A: It is known, that on $\mathbb{R}$, any monotone and bounded function converges. As for the limit you know that $f(x_{\inf}) = x_{\inf}$. Hence you can find a (potential) limit by simply solving the equation $f(x) = x$. Since you already proved that a limit exists and must lie in $(0,1)$, and limits on $\mathbb{R}$ are unique you can conclude that the solutions you obtain by solving $f(x)=x$, $x \in (0,1)$ must be "the above limit and the number the sequence converges".
A: Hints Since you studied monotonicity (derivative is positive on $(0,1)$), then $$0<x_0<1 \Rightarrow f(0)\leq f(x_0)\leq f(1) \Rightarrow 0\leq x_1\leq 1$$ and inductively, the sequence is bounded $$0\leq x_n \leq 1$$
Now, look at the function $g(x)=x-f(x)$ and see that (the detailed proof left as an exercise) $g(x)\leq 0, x\in (0,1)$ or $f(x)\geq x, x\in (0,1)$. This means that
$$x_1=f(x_0)\geq x_0 \Rightarrow x_2=f(f(x_0))\geq f(x_0)\geq x_0$$
and by induction, the sequence is monotonic. Being bounded and monotonic, means the limit exists and it is the solution for  $L=\sqrt[3]{3L^2-2L^3}$.
A: If $1 > x_n > 0$, then $x_{n+1} = x_n\left(\frac{3}{x_n}-2\right) ^{1/3} > x_n$, and thus $x_n$ is increasing. We show also that it is bounded from above, by induction. Clearly $x_0$ satisfies. Suppose $1 > x_n > 0$, then 
$$x_{n+1}^3-x_n^3=(x_{n+1}-x_n)(x_{n+1}^2+x_{n+1}x_{n}+x_n^2)=3x_{n}^2(1-x_n)$$
Thus,
$$x_{n+1}-x_n = \frac{1-x_n}{\frac{1+(x_{n+1}/x_n)+(x_{n+1}/x_n)^2}{3}}$$
Because we have already shown $x_{n+1}/x_n > 1$, the denominator is greater than 1.
Thus, $x_{n+1}-x_n<1-x_n$, or $x_{n+1} < 1$. It is also obvious that $x_{n+1}-x_n > 0$.
Therefore, we have an increasing function, bounded by above at 1. It is not much more work to show that the limit of the sequence is 1.
A: Hint: Since $f$ is continuous, if your sequence converges, the limit point $x$ is a fixed point of $f$, i.e. $f(x) = x$.
