= Solution
As $\omega\to\infty$, both <capacitors> approach <short circuits>. Thus $Z\to0$, $W\to R_v>0$, and the unloaded right junction tends to the source's top-terminal <voltage>. Hence
$$
\boxed{\lim_{\omega\to\infty}V_{\rm oc}=\frac{V_0}{2},\qquad V_A=0\ \text{for an ideal ammeter}.}
$$
The connected ideal <ammeter> carries limiting <electric current> $V_0/R$ from right to left because $Z_{\rm th}\to R/2$. Reversing the output polarity reverses both nonzero open-circuit answers, without changing their magnitudes.
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