Do we agree that the sequence $x_n$ as defined by $\begin{cases}7^\frac{1}{n} &, \impliedby n \equiv 0 \mod 2\\n^{-7} &, \impliedby n\equiv 1 \mod 2\end{cases}$ has infimum $0$ (as the second sequence goes to zero and they are both strictly decreasing), and supremum $\sqrt{7}$ (because they are both strictly decreasing and the first starts at $\sqrt{7}$ while the second starts at $1$ ? The exercice I'm doing seems to suggest the supremum is $1$. I should note that it is asking for the $\limsup$, and my problem may stem from the fact that I don't understand the difference between those if there is one.
EDIT : Actually I asked the teacher and we didn't see it in class this year, might be why I don't get that exercise. Eh, I learned something new anyway.
Ok si as far as I understand (intuitively), the limit inferior and limit superior are the same as supremum and infimum, only it is when $n\to \infty$. Hence why we don't care about the first terms and see easily that $\lim \sup =1$.