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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.

A subsequential limit of a sequence $(a_n)$ is a value $L$ approached by some subsequence $(a_{n_k})$ with strictly increasing indices $n_k$. For $a_n=(-1{)}^n(1+1/n)$, the subsequences at even integers and odd integers give subsequential limits one and minus one. A finite value appearing among only finitely many initial terms does not become a subsequential limit merely by being a member of the image of a function represented by the sequence.

For a bounded sequence of real numbers, its subsequential limits form a nonempty bounded set that is also a closed set. The Bolzano-Weierstrass theorem guarantees nonemptiness, and its extreme values are the limit inferior and limit superior. The sequence in sparse peaks can determine a limit superior has the two subsequential limits zero and one, despite almost all indices in the natural density sense carrying zero.