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Retain the exact damping factor and expand only the slowly accumulating phase. With $T=ϵt$ bounded,The resulting two-time expansion of a weakly damped linear oscillator isIn slow-time notation the second term in parentheses is $-ϵT\cos t/2$, an order-$ϵ$ correction. Simply retaining $e^{-ϵt}\sin t$ would have order-$ϵ$ rather than order-${ϵ}^2$ error on this scale, and would not meet the requested accuracy.

When ${ϵ}^{2}t=O(1)$, the phase shift is order one: Taylor-expanding its sine and cosine is no longer an ordered asymptotic expansion of the oscillatory factor. The exact solution and $z$ acquire visibly different phases after removing their common exponential damping. In contrast, for $ϵt=O(1)$ the main difference between $y$ and $z$ is their amplitude envelope: $1-ϵt$ versus $e^{-ϵt}$.

The phrase “not valid” on the later scale needs an error convention. For fixed positive damping, the absolute displacement of both $x$ and $z$ is exponentially small when ${ϵ}^{2}t$ is a positive order-one number. Indeed Taylor's theorem gives, for small positive $ϵ$,The supremum of the right side over $t\ge 0$ is $O({ϵ}^2)$. Thus $z$ loses phase accuracy on $t=O({ϵ}^{-2})$, but does not lose an absolute $O({ϵ}^2)$ error bound there. The last plot displays $e^{ϵt}x$ and $e^{ϵt}z$ to show that phase failure honestly; multiplying by the damping factor would make the difference extremely small. Relative pointwise error also becomes singular near zeros, so accuracy of the oscillatory factor is the meaningful late-time comparison.