= Solution
Let $L=\log r$. The given expansion implies $u_*=2L+O(L/r)$ and therefore
$$
e^{-u_*}=r^{-2}[1+O(L/r)].
$$
Using $H'(u)=re^{-u}(1-u)$ gives precisely
$$
\boxed{H'(u_*)=\frac1r[1-2L+O(L/r)][1+O(L/r)].}
$$
The paired low value is $v_*=H(u_*)=(2L/r)[1+O(L/r)]$. It is small, so $e^{-v_*}(1-v_*)=1+O(L/r)$ and
$$
\boxed{H'(H(u_*))=r[1+O(L/r)]\sim r.}
$$
By the <chain rule> the second-iterate <periodic-orbit multiplier> is their product:
$$
(H^2)'(u_*)=1-2L+O(L^2/r).
$$
Since $L^2/r\to0$ while $L\to\infty$, this tends to $-\infty$ and eventually has magnitude above one. Thus \b[the large second-iterate <fixed point>, and hence its associated <two-cycle>, is unstable for sufficiently large $r$]. The high state's small <derivative> alone does not imply stability: the low state's <derivative> supplies the compensating factor of order $r$.
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