Proving that $\lim_{n \to \infty}nx_n=\frac{1}{q}$ Suppose $0<x_1<1$, $x_{n+1}=x_n(1-x_n)$, prove $\lim_{n \to \infty} nx_n=1$.
Suppose now $0<x_1<\frac{1}{q}$ and  $0<q\leq 1$ and $x_{n+1}=x_n(1-qx_n)$, prove $\lim_{n \to \infty}nx_n=\frac{1}{q}$.
The first one can be done by showing the sequence is decreasing and bounded and therefore has a limit. What about the second one?
 A: Hint :
It's essentialy the same as the first part, as :
$$\lim_{n \to \infty} nx_n = \frac{1}{q} \Leftrightarrow q \lim_{n \to \infty} nx_n = 1 \Leftrightarrow \lim_{n\to \infty} nqx_n = 1 $$
But isn't that $\lim_{n\to \infty} nz_n$ = 1 where $z_n \equiv qx_n$ ?
A: 
We need first show that $\lim_{n\to\infty}n x_n=1$.  


To begin, note that for $x\in (0,1)$, $0<x(1-x)<\frac14$.  Hence, provided $x_1\in(0,1)$, $x_n>0$.  Moreover, $x_{n+1}-x_n=-x_n^2<0$. 

Inasmuch as $x_n$ is bounded below by $0$ and monotonically decreasing, the sequence converges.  Suppose it's limit is $L$.  Clearly $L=L(1-L)$.  Hence $L=0$ and we find $\lim_{n\to \infty}x_n=0$.

Let $a_n=1/x_n$.  Since $x_n\to 0$, then $a_n\to\infty$.
Hence, the Stolz-Cesaro Theorem guarantees that 
$$ \begin{align}
 \lim_{n\to\infty}\frac1{nx_n} &=\lim_{n\to\infty}\frac{a_n}n\\\\
&=\lim_{n\to\infty}(a_{n+1}-a_n)\\\\
&=\lim_{n\to\infty}\frac1{1-x_n}\\\\
&=1
\end{align}$$
as was to be shown.

For the second part, let $y_n=qx_n$ as was already mentioned in a comment and another solution.
A: We can show more than just the asymptotic behaviour
$$x_{n+1} = x_n(1-qx_n) \implies x_n = x_{n+1} + q x_{n}^2$$
divided by $x_{n+1}x_n$ results in
$$
\frac{1}{x_{n+1}} = \frac{1}{x_n} + q\underbrace{\frac{x_n}{x_{n+1}}}_{\ge 1}  \ge \frac{1}{x_n} + q
$$
Thus by the telescoping sum we have
$$
\frac{1}{x_{n}} - \frac{1}{x_1}=\sum_{k=1}^{n-1} \frac{1}{x_{k+1}} - \frac{1}{x_k} \ge (n-1)q
$$
which finally implies
$$
x_n \le \frac{1}{(n-1)q + \frac{1}{x_0}} \le \frac{1}{nq}
$$
where we have used $x_1\le 1/q$ for the last inequality (which we also need to ensure $x_{n+1}\le x_n$ by induction).
