Finding a limit of a recurrently given sequence

I am struggling with one example of limit required for recurrently given sequence. There are 3 examples in total, I will also post the ones I did.

$$x_1>0$$ and recursively given sequence: $$x_{n+1}=\frac{x_n}{1+x_n+x_n^2}$$

Find the following limits: $$\displaystyle \lim_{n \to \infty }x_n;$$ $$\displaystyle \lim_{n \to \infty }nx_n;$$ and $$\displaystyle \lim_{n \to \infty }\frac{n(1-nx_n)}{\log_{e}n}$$

First thing I do is determining whether or not the sequence is monotonically increasing or decreasing. (I already know the limit of this sequence will be $$\geq 0$$ because of $$x_1>0$$)

$$\frac{x_{n+1}}{x_n}$$ gives me $$\frac{1}{1+x_n+x_n^2}$$ which is $$<1$$ so the sequence monotonically decreases

I take some $$L= \displaystyle \lim_{n \to \infty }x_n = \displaystyle \lim_{n \to \infty }x_{n+1}$$. Swapping this into the starting sequence I get $$L=\frac{L}{1+L+L^2}$$ and from here I get $$L=0 \vee L=-1$$

Because of $$x_1>0$$, I already know $$\displaystyle \lim_{n \to \infty }x_n \neq -1$$, so $$\displaystyle \lim_{n \to \infty }x_n = 0$$ is the only solution here.

The next example is also pretty straight forward, I take $$x_n=\frac{y_n}{n}$$ and $$x_{n+1}=\frac{y_{n+1}}{n+1}$$. Here, the result is $$\displaystyle \lim_{n \to \infty }nx_n=1$$.

The third example, $$\displaystyle \lim_{n \to \infty }\frac{n(1-nx_n)}{\log_{e}n}$$, looks the trickiest and I couldn't solve it using the method I used in the 2 examples above. If anybody has any idea about solving it please share it. By the way, I am not sure if my method of solving is actually good, so if it's not please recommend/teach me a more versatile method.

Let $$y_n=\frac1{x_n}$$, then the condition $$x_{n+1}=\frac{x_n}{1+x_n+x_n^2}$$ transforms to $$y_{n+1}=y_n+1+\frac1{y_n}.$$ (1) $$y_{n+1}-y_n>1$$ implies $$\{y_n\}$$ is increasing and $$y_n=\sum_{k=1}^{n-1}(y_k-y_{k-1})>n-1$$ is unboubded above. So $$\lim_{n\to\infty}y_n=\infty\implies\lim_{n\to\infty}x_n=0.$$ (2) By Stolz's theorem, $$\lim_{n\to\infty}\frac{n}{y_n} =\lim_{n\to\infty}\frac{(n+1)-n}{y_{n+1}-y_n} =\lim_{n\to\infty}\frac{1}{1+\frac1{y_n}} =1\implies\lim_{n\to\infty}nx_n=1$$
(3) By Stolz''s theorem again (for detailed computation, see the comment below), $$\lim_{n\to\infty}\frac{y_n-n}{\log n} =\lim_{n\to\infty}\frac{\frac1{y_n}}{\log(1+1/n)} =\lim_{n\to\infty}\frac{\frac{n}{y_n}}{n\log(1+1/n)} =1.$$ Then, for the thrid limit, $$\lim_{n\to\infty}\frac{n(1-nx_n)}{\log n} =\lim_{n\to\infty}\frac{n\left(1-\frac{n}{y_n}\right)}{\log n} =\lim_{n\to\infty}\frac{n(y_n-n)}{y_n\log n} =\lim_{n\to\infty}\left[\frac{n}{y_n}\cdot\frac{y_n-n}{\log n}\right] =1.$$
• First of all, thank you for contributing! I don't understand what you did in the "First, we have, by Stolz" part? I know the Stolz-Cesaro theorem $\displaystyle \lim_{n \to \infty } \frac{a_{n+1}-a_n}{b_{n+1}-b_n}$ and then $\displaystyle \lim_{n \to \infty } \frac{a_n}{b_n}$, but I can't figure out what did you use for that limit since it's not the original example (third one is). Can you please explain a bit how you got that part? Also, I understand the first formula, but is there a more simple way to do it? Because Im not sure I would be able to think of something like that. Commented Jun 5 at 18:57
• $$\lim_{n\to\infty}\frac{y_n-n}{\log n}=\lim_{n\to\infty}\frac{(y_{n+1}-(n+1))-(y_n-n)}{\log(n+1)-\log n}=\lim_{n\to\infty}\frac{y_{n+1}-y_n-1}{\log(1+1/n)}=\cdots.$$ Commented Jun 5 at 23:59
• The key point is $$y_{n+1}=y_n+1+\frac1{y_n},$$ where $y_n=1/x_n$. Commented Jun 6 at 0:07
• Also you need know $$\lim_{n\to\infty}n\log(1+1/n)=1.$$ Commented Jun 6 at 0:23