Sequential limit.

First, we have $a_{n+1} -a_n=-a_n^2<0$ and $a_n>0 \Leftrightarrow a_{n+1}>0$ , Which means that $(a_n)$ is positively decreasing and therefore convergent and the limit is clearly $0$ .

We have $\frac{1}{a_{n+1}} - \frac{1} {a_{n}} = \frac{1}{1-a_n} \to 1$ Then by Cesaro-Stolz theorem , we get $na_n \to 1$ .

We have $\lim_{n\to \infty}\frac{n(1-n a_n)}{\ln n} = \lim_{n\to \infty} \frac{\frac{1}{a_n}-n}{\ln n}$ And we have : $\frac{\frac{1}{a_{n+1}} - \frac{1}{a_n} -1}{\ln(n+1) -\ln n } = n a_n \cdot \frac{1}{1-a_n} \cdot \frac{1}{n\ln\left(1+\frac{1}{n} \right)} = 1.$ Then (by Cesaro Stolz) the result is $\boxed{1}$ .