What is $\lim_{n \to \infty} n a_n$? Let $L(x) = x - \frac{x^2}{2}$ and let 
$$ a_n = \underbrace{L(L(\cdots L(}_{n \text{ times}}\frac{17}{n}))\cdots).$$
I'm trying to find $\lim_{n \to \infty} n a_n$. 
I don't really know how to proceed. I tried calculating the first few values of $na_n$ and plotted them and got 

The limit seems to be somewhat above $1.6$. (By the way, I'm not sure why the horizontal axis is labeled like that. I plotted $na_n$ from for $n \in \{ 10, \cdots, 23 \}$, because the early values of $n a_n$ blow up. But for some reason the labeling started from $1$.)
How could I go about finding the limit? There has to be some sort of trick (this is apparently from the high school math competition "Math Prize for Girls", but I found this problem on expii), but I'm just not seeing it...
Edit: Frenzy Li linked to this question. The question is the same, but the answer given back then (2) appears to be incorrect. (Based on computing large values, it appears to be $\frac{34}{19}$.)
 A: To compute $na_n$ we start with $y=17$  and apply $n$ times the transformation
$$y_{new}=nL(y/n)=y-y^2/(2n)$$
i.e.
$$(y_{new}-y)/\epsilon=-y^2/2$$
where $\epsilon=1/n$.
For $n\to \infty$ this means that we're solving the ODE
$$y'=-y^2/2$$
with the initial condition $y(0)=17$, and want to know $y(1)$. The solutions of the ODE are $y(x)=2/(x+c)$, $y(0)=17$ gives $c=2/17$, hence $y(1)=34/19$.
edit: In more detail: for a given $n$ we define a function 
$$y_n:\{0,\frac1n,\frac2n,\frac3n,\dots,1\}\to \mathbb R$$
via
$$y_n(0)=17,\quad y\bigl(x+\frac1n\bigr)=y(x)-\frac{y(x)^2}{2n}$$
and then extend $y_n$ to a piecewise-linear function $[0,1]\to \mathbb R$. We have $y_n(1)=na_n$. This function $y_n$ is also the outcome of the Euler approximate method of solving the ODE $y'=-y^2/2$ with the initial condition $y(0)=17$ and with the step $1/n$. The sequence $y_n$ converges to a solution of the ODE - that's true for any ODE whenever the RHS is (say) locally Lipschitz in $y$. Not sure about a textbook reference, here is what I found on MO: https://mathoverflow.net/a/200412/9390
A: @user8268's answer is amazing. I was shocked to see how ideas derived from numerical method could benefit problems that are totally unrelated at first glance. Just want to add a few things.
As mentioned in the answer. The ODE discretization is in $[0, 1]$ with resolution $1/n$. The figure below shows the dsicretization with $n = 8, 16, 32, 64$. $n=8$ corresponds to the bottom curve, $n=16$ corresponds to the second to bottom curve, and so on. The convergence is apparent.

Another interesting observation is the solution "blows up" numerically for $n = 7$(not shown), and becomes negative for $n = 8$. Those erroneous behaviors is very common in numerical solutions of ODE where the resolution is insufficient. Here we can estimate the minimum resolution required.
The discretization is 
$$
y^{i+1} = y^i \left( 1 - \frac{y^i}{2n} \right)
$$
To maintain the positive definiteness, we must have $n > y^i/2$. Apparent $y$ is decreasing according to the ODE, so $\max(y) = y^0 = 17$, then we have $n > 17/2$ or $n = 9$.
