jmc

Note that $0<L(x)<x$ for $0<x<2.$ Assuming $n$ is sufficiently large, i.e. $n\ge 9,$ we have that $0<a_n<\frac{17}{n}<2.$

From $L(x)=x-\frac{x^2}{2},$ we can write $$\frac{1}{L(x)}=\frac{1}{x-\frac{x^2}{2}}=\frac{2}{2x-x^2}=\frac{2}{x(2-x)}=\frac{x+(2-x)}{x(2-x)}=\frac{1}{x}+\frac{1}{2-x},$$so $$\frac{1}{L(x)}-\frac{1}{x}=\frac{1}{2-x}(\ast ).$$For a nonnegative integer $k,$ let $L^{(k)}(x)$ denote the $k$th iterate of $L(x).$ Then $0<L^{(k)}(x)<x,$ so $$0<L^{(k)}\left(\frac{17}{n}\right)\le \frac{17}{n}.$$Hence, $$\frac{1}{2}<\frac{1}{2-L^{(k)}(\frac{17}{n})}\le \frac{1}{2-\frac{17}{n}}=\frac{n}{2n-17}.$$By equation $(\ast ),$ $$\frac{1}{L^{(k+1)}(\frac{17}{n})}-\frac{1}{L^{(k)}(\frac{17}{n})}=\frac{1}{2-L^{(k)}(\frac{17}{n})},$$so $$\frac{1}{2}<\frac{1}{L^{(k+1)}(\frac{17}{n})}-\frac{1}{L^{(k)}(\frac{17}{n})}\le \frac{n}{2n-17}.$$Summing over $0\le k\le n-1,$ we get $$\frac{n}{2}<\frac{1}{L^{(n)}(\frac{17}{n})}-\frac{1}{\frac{17}{n}}\le \frac{n^2}{2n-17}.$$Since $a_n=L^{(n)}\left(\frac{17}{n}\right),$ this becomes $$\frac{n}{2}<\frac{1}{a_n}-\frac{n}{17}\le \frac{n^2}{2n-17}.$$Dividing by $n,$ we get $$\frac{1}{2}<\frac{1}{n{a}_n}-\frac{1}{17}\le \frac{n}{2n-17}.$$As $n$ approaches infinity, $\frac{n}{2n-17}$ approaches $\frac{1}{2},$ so if $L$ is the limit of $n{a}_n,$ then $$\frac{1}{L}-\frac{1}{17}=\frac{1}{2}.$$Solving, we find $L=\boxed{\frac{34}{19}}.$