Is my proposition correct?

Question

I have concluded (proved) the following, but haven’t seen it stated anywhere, hence would like to get it verified by you mathematicians.

Let $(a_n)_{n=m}^\infty$ be a sequence of reals for some $m\in\mathbb{Z}$ such that $(|a_n|)_{n=m}^\infty$ converges to some $L\in\mathbb{R}$. (Hence, $L\ge 0$.) Then $L$ or $-L$ (or both) is a limit point of $(a_n)_{n=m}^\infty$. Further, if $L$ is a limit point, then $L$ is the limit superior of $(a_n)_{n=m}^\infty$ and if $-L$ is a limit point, then $-L$ is the limit inferior of $(a_n)_{n=m}^\infty$.

Answer 1

(Yes it's true) Hint : At least one of the two sets $A_+ := \{n : a_n > 0\}$ and $A_- := \{n : a_n \leq 0\}$ is infinite. W.L.O.G suppose that $A_+$ is infinite; then it immediately follows that $L$ is a limit point of $(a_n)_n$. And therefore $$\varlimsup (a_n)_n = \varlimsup_{n \in A_+} (a_n)_n = \lim_{n \in A_+}(a_n)_n = L. $$

Answer 2

You are correct.

An easy way to see this is by the following characterization of a limit:

$$ \lim_{n\rightarrow \infty}x_n = L \quad \iff \quad \liminf_{n \rightarrow \infty}{x_n} = \limsup_{n \rightarrow \infty}{x_n} = L $$

In particular, if $|a_n| \rightarrow L$, then $\limsup{|a_n|} = L$. But (it is easy to check that) this is the case iff either $\limsup{a_n} = L$ or $\liminf{a_n} = -L$.