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?