I came across an exercise (Exercise 10, Ch. 1, Marsden's elementary classical analysis, 2nd ed.) that gives a characterization of lim sup I had never seen, which can be rephrased as follows:
Let $(x_{n})$ be a sequence in $\mathbb{R}$. Then there is some $l \in \mathbb{R}$ such that $\limsup x_{n} = l$ if and only if for every $\varepsilon > 0$ there is some $N \geq 1$ such that $x_{n} < l+\varepsilon$ for all $n \geq N$ and $x_{n} > l-\varepsilon$ for some $n \geq N$.
The "only if" part is obvious, but I am afraid (not sure) the "if" part is false. For, let $(x_{n})$ be the sequence such that $x_{1} := 1$ and $x_{n} := 0$ for all $n \geq 2$. Then for every $\varepsilon > 0$ we have $x_{n} < 1+\varepsilon$ for all $n \geq 1$ and, since $x_{1} = 1$, we have $x_{n} > 1 - \varepsilon$ for some $n \geq 1$. But $\limsup x_{n} = 0 \neq 1$. Is my counterexample wrong or is the exercise wrong?
