= Solution
Put $\theta=k\log t$. Since
$$
\dot x=\frac{\sin\theta+2k\cos\theta}{2k\sqrt t},
\qquad
\epsilon^{-2}=k^2+\frac14,
$$
the instantaneous oscillator energy is
$$
E(t)=\frac12\dot x^2+\frac{x^2}{2(\epsilon t)^2}
=\frac{1-\cos2\theta+2k\sin2\theta+4k^2}{8k^2t}.
$$
The instantaneous frequency is $\omega(t)=1/(\epsilon t)$, so
$$
I(t)=\frac{E(t)}{\omega(t)}
=\frac{\epsilon}{8k^2}\left(4k^2+1-\cos2\theta+2k\sin2\theta\right).
$$
Its oscillatory part divided by its mean is $O(k^{-1})=O(\epsilon)$. It remains bounded and periodic in $\log t$, so the variations do not accumulate. This is the expected behavior of an <adiabatic invariance of the action>[adiabatic invariant]: slow parameter variation produces small bounded oscillations rather than secular drift.
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