Q.For a very high frequency AC supply, a capacitor behaves like a pure conductor. Why?

[!TLDR] The capacitive reactance $X_C=\dfrac{1}{2\pi fC}$ tends to zero as f becomes very large, so a capacitor offers negligible opposition to very-high-frequency AC, effectively behaving like a plain conducto …

The capacitive reactance of a capacitor is $X_C=\dfrac{1}{2\pi fC}$, which is INVERSELY proportional to the frequency f of the applied AC. As f is made larger and larger (a 'very high frequency' supply), $X_C$ correspondingly becomes smaller and smaller, approaching zero in the limit of very high f. Since $X_C$ is the effective opposition the capacitor offers to the flow of current, a vanishingly small $X_C$ means the capacitor offers almost no resistance at all to the current -- it lets the AC pass through essentially unimpeded, exactly as an ideal, resistance-free conducting wire would. This is the opposite extreme to the DC case ($f=0$), where $X_C\to\inft …

Examine how $X_C=1/(2\pi fC)$ behaves in the limit of large f, and connect a vanishing reactance to negl …

Confusing the high-frequency limit of a capacitor ($X_C\to0$, behaves like a conductor) with the high-frequency limit of an inductor ($X_L\to\infty$, behaves like an open circuit …