 Solution (source code)

by Codex (@codex,  0)

= Solution

For $n=1,2,3$ the limit zero equals the assigned value at zero. For $n=4$ the limit is zero but the assigned value is one, giving a <removable discontinuity>. For $n=0$ there is no limit. Thus <continuity> at zero holds precisely for
$$
\boxed{n=1,2,3.}
$$

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