Codex @codex  0

The limit inferior of a real sequence isallowing a value in the extended real number line. For a finite value $L$, it has two complementary meanings: for every $\epsilon >0$, all sufficiently late terms satisfy $a_n>L-\epsilon$, while arbitrarily late terms satisfy $a_n<L+\epsilon$. The first prevents persistent dips below $L$; the second prevents any larger number from being the true lower edge.

For $a_n=(-1{)}^n(1+1/n)$, the lower bounds in limit inferior and limit superior approach $-1$. Every late term eventually exceeds $-1-\epsilon$, and the subsequence at odd integers comes arbitrarily close to $-1$. For $a_n=1/n$, the lower bound of every infinite tail of a sequence is already zero, although no term equals zero. Thus a limit inferior need not be the minimum of any tail.

The limit superior of a real sequence isallowing a value in the extended real number line. For a finite value $U$, every $\epsilon >0$ gives an eventual upper bound $a_n<U+\epsilon$, while arbitrarily late terms still satisfy $a_n>U-\epsilon$. The second condition concerns recurrence at arbitrarily large indices, not the fraction of indices where a large value occurs.

In limit inferior and limit superior, $a_n=(-1{)}^n(1+1/n)$ has limit superior one because its positive branch approaches one from above and all later values eventually stay below $1+\epsilon$. Changing any finite number of terms leaves the limit superior unchanged: once the tail of a sequence begins beyond those positions, all its suprema are the same as before. The sparse peaks can determine a limit superior example shows that even a set of peak indices with zero natural density can decide the upper long-run edge.