If $x_{n+1} = (1+\frac{x_n}{a})^a - 1$, find $\lim_{n\to\infty}nx_n$ If $0 < a < 1, x_1>0$ and $x_{n+1} = \left(1+\frac{x_n}{a}\right)^a-1$ compute $\lim\limits_{n\to \infty} nx_n$.
I proved that all $x_n$ are positive (with induction) and $x_n$ is decreasing (with Bernoulli), so convergent. Now, if I take the limit, I get:
$l=\left(1+\frac{l}{a}\right)^a-1$
with $l=\lim\limits_{n\to \infty}x_n$, but I don't know how to solve this or how to continue after this. I appreciate any help.
 A: So $l=\lim\limits_{n\to \infty} x_n \in [0,x_1)$ and you get
$$l=\left(1+\frac{l}{a}\right)^a-1$$
Now, differentiating with respect to $l$ gives
$$1=a\cdot\left(1+\frac la\right)^{a-1}\cdot\frac 1a$$
and from here it's easy to deduce that $l = 0$. To evaluate your limit, we can use Cesaro-Stolz and l'Hopital:
$$
\begin{aligned}
\lim_{n\to\infty} nx_n &= \lim_{n\to \infty} \frac{n}{\frac{1}{x_n}}\\
&= \lim_{n\to\infty}\, \frac {(n+1)-n}{\frac 1{x_{n+1}}-\frac 1{x_n}}\\
&= \lim_{n\to\infty}\, \frac {x_nx_{n+1}}{x_n-x_{n+1}} \\
&= \lim_{n\to\infty}\, \frac {x_n\left[\left(1+\frac {x_n}a\right)^a-1\right]}{x_n-\left(1+\frac {x_n}a\right)^a+1}\\
&= \lim_{n\to\infty}\left[ \frac {\left( 1+\frac {x_n}a\right )^{a-1}}{\frac {x_n}a\cdot\frac {x_n^2}a\cdot\frac 1{1+x_n}-\left(1+\frac {x_n}a\right)^a} \right]\\
&= \lim_{n\to\infty}\, \frac {x_n^2}{1+x_n-\left(1+\frac {x_n}a\right)^a}\\
&= \lim_{x\to 0}\, \frac {x^2}{1+x-\left(1+\frac xa\right)^a}\\
&= \lim_{x\to 0}\, \frac {2x}{1-a\cdot\left(1+\frac xa\right)^{a-1}\cdot\frac 1a}\\
&= 2\lim_{x\to 0}\, \frac x{1-\left(1+\frac xa\right)^{a-1}}\\
&= 2\lim_{x\to 0}\, \frac 1{\left(1-a\right)\cdot\left(1+\frac xa\right)^{a-2}\cdot\frac 1a} \\
&=\boxed{\frac{2a}{1-a}}
\end{aligned}
$$
Note: When using a Cesaro-Stolz, for the fluency of the proof, I (almost always) write:
$$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n\to \infty} \frac{a_{n+1}-a_{n}}{b_{n+1}-b_n}$$
However this is a notation abuse. Correct it is to prove first that the limit 
$$\lim_{n\to \infty} \frac{a_{n+1}-a_{n}}{b_{n+1}-b_n}$$
exists and is finite (say it equals $l$), and then say
$$\lim_{n \to \infty} \frac{a_n}{b_n} =l$$
