Given $g:\mathbb{R}\to\mathbb{R}$, we have that $\lim\limits_{x\to\infty}g(x)=a$ if, and only if, the sequence of functions $f_n(x)=g(x+n)$ converges uniformly in $[0,\infty)$ to the constant function $f(x)=a$.
My work so far:
$(\Rightarrow)$ The definition of limit to $\lim\limits_{x\to\infty}g(x)=a$: $$\forall~\varepsilon>0,\exists K>0\mbox{ such that }x>K\Rightarrow|g(x)-a|<\varepsilon.$$
So, from this, we got: $$|g(x)-a|<\varepsilon\Rightarrow|g(x+n)-a|<\varepsilon\Rightarrow|f_n(x)-a|<\varepsilon,$$ in other words, $\lim\limits_{x\to\infty}f_n(x)=a=f(x),\forall~x\in[0,\infty)$, soon, $f_n$ converges uniformly to $f(x)=a$.
Comment: I tried to use the fact that the limit is for all $x\in[0,\infty)$ and just switch the $x$ to $x+n$, like $x=x+n$ or something, but I don't know if this works and it's valid. Further, I think that I need to have something with $n>n_0\in\mathbb{N}$ to ensure the convergence $|f_n(x)-f(x)|<\varepsilon$.
$(\Leftarrow)$ Now, $f_n$ converges uniformly to $f(x)=a$, in other words, for all $\varepsilon>0$ exists $n_0\in\mathbb{N}$ such that $n>n_0(\varepsilon)$ implies that $$|f_n(x)-f(x)|<\varepsilon,\forall~x\in X$$
Like this, $$|f_n(x)-f(x)|<\varepsilon\Rightarrow|g(x+n)-a|<\varepsilon\Rightarrow$$
$$\Rightarrow|g(x)-a|<\varepsilon,\forall~x\in X,$$
therefore, $\lim\limits_{x\to\infty}g(x)=a$.
Comment: My questions here are pretty similar to my previously comment. I used the same ideia that it's for all $x\in X$, then it's valid to $x+n=``k"=x$, but I don't know if it's really valid. Beside that, I think that I need that $x>K$ implies that $|g(x)-a|<\varepsilon$ and I don't know how to have that.
Every help is welcome. Thanks :D
In general: If you doubt any of your arguments, try to include more intermediate steps.
Let us start with $\Rightarrow$:
Assume $\varepsilon>0$. We need to find $n_0\in \mathbb{N}$ such that for all $n\in \mathbb{N}$ with $n\geq n_0$ and all $x\in[0, \infty)$ it holds $\vert f_n(x)-f(x)\vert=\vert f_n(x)-a\vert<\varepsilon.$ We know there exists $K>0$ with $\vert g(\tilde{x})-a\vert <\varepsilon$ for all $\tilde{x}>K$. Now set $n_0=\lceil K \rceil+1$. Then for $n\geq n_0$ and $x\in[0, \infty)$ we have $x+n\geq x+n_0>K$ and therefore $\vert f_n(x)-a\vert=\vert g(n+x)-a\vert<\varepsilon$.
I'll let you try to improve the other direction.