Find limit of $\frac{n(nx_n-\frac{1}{3})}{\ln n}$ knowing that $x_{n+1}=x_n-3x_n^2$ Let $x_n$ a sequence with $\frac{1}{3}>x_1>0$ and $x_{n+1}=x_n-3x_n^2$. Find the limit 
$$\lim_{n\to \infty} \frac{n(nx_n-\frac{1}{3})}{\ln n}.$$
I proved that $x_n$ is convergent to $0$ and that $\lim\limits_{n\to \infty} nx_n= \frac{1}{3}$ with Stolz-Cesaro theorem.
$$\lim_{n\to \infty} \frac{n}{\frac{1}{x_n}}=\lim_{n\to \infty} \frac{1}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=\lim_{n\to \infty} \frac{x_{n+1}x_n}{x_n-x_{n+1}}=\lim_{n\to \infty}\frac{x_n(x_n-3x_n^2)}{3x_n^2}=\frac{1}{3}$$
I also know that $\lim\limits_{n\to \infty}\frac{n}{\ln n}=\infty$. The limit is $\infty \cdot 0$ and I am stuck.
Can I get a clue or a hint please?

Firstly,
$$\lim_{n\to \infty} \frac{n\left(nx_n-\frac{1}{3}\right)}{\ln n}=\lim_{n\to \infty} nx_n \cdot \lim_{n\to \infty} \frac{n-\frac{1}{3x_n}}{\ln n}=\frac{1}{3} \lim_{n\to \infty}\frac{n-\frac{1}{3x_n}}{\ln n}$$
Now, from Cesaro-Stolz:
$$
\begin{aligned}
\lim_{n\to \infty}\frac{n-\frac{1}{3x_n}}{\ln n} &= \lim_{n\to \infty} \frac{n+1-\frac{1}{3x_{n+1}}-n+\frac{1}{3x_n}}{\ln(n+1)-\ln n}\\
&= \lim_{n\to \infty} \frac{1+\frac{1}{3x_n}-\frac{1}{3x_{n+1}}}{\ln\left(\frac{n+1}{n}\right)}\\
&=\lim_{n\to \infty} \frac{n\left(1+\frac{x_{n+1}-x_n}{3x_nx_{n+1}}\right)}{\ln\left(\frac{n+1}{n}\right)^n}\\
&= \lim_{n\to \infty} n\left(1-\frac{x_n}{x_{n+1}}\right)\\
&= \lim_{n\to \infty} \frac{n(x_{n+1}-x_n)}{x_{n+1}}\\
&= \lim_{n\to \infty} \frac{n(-3x_n^2)}{x_n-3x_n^2}\\
&= -\lim_{n\to \infty} nx_n \cdot \lim_{n\to \infty} \frac{3}{1-3x_n}\\
&=-1
\end{aligned}
$$
Thus, the final result is:
$$\lim_{n\to \infty} \frac{n\left(nx_n-\frac{1}{3}\right)}{\ln n}=-\frac{1}{3}$$ 
