Can limit of hard languages be easy?
I think we can just start with some base language $L$, then take $L_0 = L$ and $L_{s+1} = L_s \cup \{0,1\}^{s+1}$.

That is, each $L_s$ is the union of $L$ with all strings of length up to $s$. Each $L_s$ is at least as hard as $L$ but is no harder (in an asymptotic sense), assuming we can count to $s$.

I also thought about the opposite "limit", so each $L_{s+1}$ is contained in $L_s$, and $L = \cap_s L_s$ is easy while each $L_s$ is hard. But I think we could just start with a hard (but countable) language $L_0$ and just remove one word at each step; the intersection should be empty (every word is eventually removed).