I was wondering if the following statement is true, and if my proof works. I also wonder if the statement is true if I allow $a=\pm\infty$?
Statement: Let $a\in\mathbb{R}$ and $f:\mathbb{R}\to\mathbb{R}$. Then $$a=\limsup_{r\to\infty}f(r)\quad\Leftrightarrow\quad \forall\varepsilon>0: a-\varepsilon\stackrel{n}{<}f(r)\stackrel{as}{<}a+\varepsilon$$ (where $\stackrel{n}{<}$ means that the inequality holds for a sequence $(r_n)$ tending to infinity and $\stackrel{as}{<}$ means that the inequality holds asymptotically, i.e. for all $r>r_0$).
Proof: ''$\Rightarrow$'' Let $\varepsilon>0$ be given. Then there exists $R_0$ such that for $R\geq R_0$: $$ 0\leq\sup_{r\geq R}f(r)-a\leq\varepsilon/2 $$ implying $f(r)\stackrel{as}{<}a+\varepsilon$.
Assume (for contradiction) that there does not exist a sequence $(r_n)$ tending to infinity and satisfying $a-\varepsilon<f(r_n)$ for all $n$. Then there must exist $R_0$ such that $a-\varepsilon\geq f(r)$ for al $r\geq R_0$. But then $$ a-\varepsilon\geq\sup_{r\geq R_0}f(r)\geq\limsup_{r\to\infty}f(r)=a. $$ A contradiction.
''$\Leftarrow$'' Since for all $\varepsilon>0$, $f(r)\stackrel{as}{<}a+\varepsilon$, we have that $$ \forall\varepsilon>0\:\exists R_{\varepsilon}:\quad a+\varepsilon\geq\sup_{r\geq R_\varepsilon}f(r)\geq\limsup_{r\to\infty}f(r), $$ proving that $a\geq\limsup_{r\to\infty}f(r)$.
Since for all $\varepsilon>0$, $a-\varepsilon\stackrel{n}{<}f(r)$, we have that $$ \forall\varepsilon>0\:\forall R\in\mathbb{R}\:\exists r>R:\quad a-\varepsilon<f(r). $$ Hence $$ \forall\varepsilon>0\:\forall R\in\mathbb{R}:\quad a-\varepsilon<\sup_{r\geq R}f(r), $$ which implies $$ \forall\varepsilon>0:\quad a-\varepsilon\leq\limsup_{r\to\infty}f(r), $$ proving that $a\leq\limsup_{r\to\infty}f(r)$.