Finding $\displaystyle \lim_{n\to \infty} (x_0 x_1 \cdots x_n)\sqrt{n}$ where $x_{n+1}=x_n^3-x_n^2+1$, $x_0=\frac{1}{2}$ 
Problem. Let $(x_n)$ be the sequence defined by $x_0=\frac{1}{2}$ and $x_{n+1}=x_n^3-x_n^2+1$ for any $n\in \mathbb{N}\cup \{0\}$. Find
$$ \lim_{n\to \infty} (x_0 x_1 \cdots x_n)\sqrt{n}.$$

According to the answer sheet, this limit equals $1$. However, I can't manage to solve it. Here is what I've done.
Obviously, $x_{n+1}-x_n=(x_n-1)^2(x_n+1)>0$ (it is easy to observe that all the terms of the sequence are positive), so $(x_n)$ is a strictly increasing sequence.
Let us now prove by induction on $n$ that $x_n<1$ for all $n\in \mathbb{N}\cup \{0\}$. The base case is obvious, so suppose it holds for $n$ and prove it for $n+1$. $x_{n+1}=x_n^2(x_n-1)+1<1$ by the induction hypothesis and we are done.
Hence, $(x_n)$ is monotone and bounded, so it is convergent. It is easy to see now that $\displaystyle \lim_{n\to \infty}x_n=1$.
Now I pretty much got stuck. I tried to use the epsilon definition of a limit, trying to exploit $\displaystyle \lim_{n\to \infty}x_n=1$, but it didn't help. Maybe I should use Stolz–Cesàro on the limit that I want to compute?
 A: This question is similar to previous MSE questions including
question 2675217,
question 2776443,
question 2861768.
Define the recursion
$$ a_{n+1}=f(a_n)\;\;\text{ where }\;\; f(x):=x(1-x)^2. \tag{1}$$
Define the sequence $\, x_n := 1-a_n.\,$ Verify that
$\, x_{n+1}=x_n^3-x_n^2+1.\,$
Assume an asymptotic expansion
$$a_n = \sum_{k=0}^\infty  p_k(c,z)y^{k+1},
\;\; y := 1/n,\quad z := \log(y) \tag{2}$$
where $\,p_k(c,z)\,$ is a
polynomial of degree $\,k\,$ in $\,c\,$ and $\,z.\,$
Substitute equation $(2)$ into equation $(1)$ and
solve for the polynomials $\,p_k\,$ to get
$$ p_0\!=\!\frac12,\, p_1\!=\!\frac{c\!+\!3z}8,\,
p_2\!=\!\frac{(5\!+\!3c\!+\!c^2)\!+\!
(9\!+\!6c)z\!+\!9z^2}{32}. \tag{3} $$
Define the sequence
$$y_n:=(x_0x_1...x_n)\sqrt{n}. \tag{4} $$
Use equation $(1)$ and the definition of $\,x_n\,$ to verify that
$$ y_n=\sqrt{na_{n+1}/a_0}. \tag{5}$$
Use equation $(2)$ to get
$$\lim_{n\to\infty} y_n = \sqrt{\frac{p_0}{a_0}} = \frac1{\sqrt{2a_0}}. \tag{6}$$
When $\,x_0=a_0=\frac12,\,$ the limit is $1$. Also, $\,c\approx -8.205896246.$
