Let $(a_{n})_{n=m}^{\infty}$ be a sequence of real numbers, let $L^{+}$ be the limit superior of this sequence, and let $L^{-}$ be the limit inferior of this sequence (thus both $L^{+}$ and $L^{-}$ are extended real numbers).
Proposition
For every $x > L^{+}$, there exists an $N\geq m$ such that $a_{n} < x$ for all $n\geq N$. (In other words, for every $x > L^{+}$, the elements of the sequence $(a_{n})_{n=m}^{\infty}$ are eventually less than $x$). Similarly, for every $y < L^{-}$ there exists an $N\geq m$ such that $a_{n} > y$ for all $n\geq N$.
Proof
According to the definition of limit superior, one has that \begin{align*} L^{+} = \limsup_{n\rightarrow\infty}a_{n} = \inf(a^{+}_{M})_{M=m}^{\infty} \end{align*} where $a^{+}_{M}$ is the real number defined by $a^{+}_{M} = \sup\{a_{n}: n\geq M\}$.
Consequently, if $x > L^{+}$, there is a natural number $N$ such that $L^{+} \leq a^{+}_{N} < x$.
But, accordingly to the corresponding definition, we have that $x > \sup\{a_{n}: n\geq N\}$, that is to say, $x > a_{n}$ for $n\geq N$.
On the other hand, according to the definition of limit inferior, we have that \begin{align*} L^{-} = \liminf_{n\rightarrow\infty}a_{n} = \sup(a^{-}_{N})_{N=m}^{\infty} \end{align*} where $a^{-}_{M}$ is the real number defined by $a^{-}_{M} = \inf\{a_{n}:n\geq M\}$.
Hence, if $y < L^{-}$, there is a natural number $N$ such that $y < a^{-}_{N} \leq L^{+}$.
But, accordingly to the corresponding definition, we have that $y < \inf\{a_{n}: n\geq N\}$, that is to say, $a_{n} > y$ for $n\geq N$.
Could someone please verify if my proof is rigorously correct?