Show that if $\lim\limits_{x\to+\infty}f(x)$ exists and if $f(x_n)\to\eta$ for some sequence $(x_n)$ with $x_n \to+\infty$, so $\lim\limits_{x\to+\infty}f(x)=\eta$.
Hence $\lim\limits_{x\to+\infty}f(x)$ exists, $f$ has to be defined on an interval $I$ which is unbound on the right side.
The existence of $\lim\limits_{x\to+\infty}f(x)$ also implies it approaches a real number, we call $\xi$. So $$\lim\limits_{x\to+\infty}f(x)=\xi.$$
The existence of the limit at $\xi$ is equivalent to
$$\forall \epsilon>0 \,\,\, \exists \vartheta \in I \,\,\,\forall x> \vartheta: |f(x)-\xi|< \epsilon. $$
Since $x_n \to+ \infty$ we know that $\forall\vartheta\,\,\,\exists N \in \mathbb{N} \forall m> N:x_m> \vartheta$
This means nothing else than:
$$\forall \epsilon>0 \,\,\, \exists \vartheta \in I \,\,\,\exists N \in \mathbb{N} \,\,\,\forall m>N: |f(x_m)-\xi|< \epsilon,$$
which means $f(x_m)\to\xi$. But since $(f(x_m))$ is a subsequence of $(f(x_n))$ for which $f(x_n) \to\eta$ and the limit of a converging sequence is explicit, $\xi=\eta$, which means:
$$\forall \epsilon>0 \,\,\, \exists \vartheta \in I \,\,\,\forall x> \vartheta: |f(x)-\eta|< \epsilon,$$ which says $$\lim\limits_{x\to+\infty}f(x)=\eta.$$
It would be great if someone could look over it and say if my reasoning is correct, and if not, what I could improve.