Show $\lim_{n\to\infty}nx_n=1$ without using Stolz. 
Let $x_0\in (0,1)$, $x_{n+1}=x_n(1-x_n)\ (n\geq 0)$. Then it is easy to show $x_n\to 0$, $\forall\ n\geq 1, (n+1)x_n<1$; $nx_n\nearrow, n\geq 1$. But how can we show $\lim_{n\to\infty}nx_n=1$ without using Stolz?

If we use Stolz, it is easy. $\lim nx_n=\lim \frac{n}{x_n^{-1}}=\lim \frac{1}{x_n^{-1}-x_{n-1}^{-1}}=\lim (1-x_{n-1})=1.$ But how can we prove it withou using Stolz?
 A: You have
$$\begin{aligned}
\frac{1}{x_{n+1}} - \frac{1}{x_n} &= \frac{1}{x_n(1-x_n)} - \frac{1}{x_n}\\
&=\frac{1}{1-x_n}  \underset{n \to \infty}{\to} 1
\end{aligned}$$
Hence for any $\epsilon \gt 0$, it exists $N \in \mathbb N$, such that for $n \ge N$
$$1-\epsilon \le \frac{1}{x_{n+1}} - \frac{1}{x_n} \le 1+ \epsilon$$ and suming up those inequalities
$$(n-N)(1- \epsilon) \le \frac{1}{x_{n}} - \frac{1}{x_N} \le (n-N)(1+ \epsilon)$$ which is equivalent to
$$\left(1 - \frac{N}{n}\right)(1-\epsilon)+\frac{1}{nx_N} \le \frac{1}{nx_n} \le \left(1 - \frac{N}{n}\right)(1+\epsilon)+\frac{1}{nx_N}$$ and proves the desired result as for $n \to \infty$, the LHS converges to $1-\epsilon$ and the RHS to $1+\epsilon$.
A: This answer essentially includes a proof of Stolz-Cesaro, but does not cite Stolz-Cesaro.

Let $0\lt x_0\lt1$ and
$$
x_{n+1}=x_n(1-x_n)\tag1
$$
Next, set $a_n=\frac1{x_n}$; then
$$
a_{n+1}=\frac1{x_n(1-x_n)}=\frac{a_n^2}{a_n-1}\tag2
$$
Cross multiplication gives
$$
a_n+1\le\overbrace{\ \frac{a_n^2}{a_n-1}\ }^{a_{n+1}}\le a_n+1+\frac2{a_n}\tag3
$$
Subtracting $a_n$ from $(3)$ yields
$$
1\le a_{n+1}-a_n\le1+\frac2{a_n}\tag4
$$
The left-hand inequality in $(4)$ guarantees that $a_n\ge a_0+n$. Thus,
$$
1\le a_{n+1}-a_n\le1+\frac2n\tag5
$$
For $n\gt m$,
$$
1\le\frac{a_n-a_m}{n-m}\le1+\frac2m\tag6
$$

Taking the $\limsup$ and $\liminf$ of $(6)$, and noting that $\limsup\limits_{n\to\infty}\frac{a_n}n=\limsup\limits_{n\to\infty}\frac{a_n-a_m}{n-m}$ and $\liminf\limits_{n\to\infty}\frac{a_n}n=\liminf\limits_{n\to\infty}\frac{a_n-a_m}{n-m}$, gives
$$
1\le\liminf_{n\to\infty}\frac{a_n}n\le\limsup_{n\to\infty}\frac{a_n}n\le1+\frac2m\tag7
$$
for arbitrary $m$.  Therefore,
$$
\lim_{n\to\infty}\frac{a_n}n=1\tag8
$$
and thus,
$$
\lim_{n\to\infty}nx_n=\lim_{n\to\infty}\frac{n}{a_n}=1\tag9
$$
A: The conclusion can be derived by a tool very similar in spirit to the Stolz theorem. We have $$y_n:=x_{n}^{-1}-x_{n-1}^{-1}={1\over 1-x_{n-1}}\to 1$$ Hence for $y_1=x_1^{-1}$ we get
$${1\over nx_n}={1\over n}[y_1+y_2+\ldots +y_n]\to 1$$
