Prove that $f\colon [a;\infty) \to \mathbb{R} $ is uniform continuous if $f$ is continuous. We have the continuous function $f\colon [a;\infty) \to \mathbb{R}$ .
 How can i prove that it is uniformly continuous?
 The function has a finite limit in $+\infty$.
I know that for a pair of $a,b \in \mathbb{R}$, $f\colon[a;b) \to\mathbb{R}$ is continuous if the $ \lim\limits_{\substack {x\to b \\x <b} }  f(x) $ is finite.
 I have absolutely no idea how to prove it. Can someone offer a step by step proof or an article to the proof?

Following the exchange in the comments, posting what what there as answer.


*

*This is false without further assumptions, as the example of $f\colon x\in[a,\infty)\mapsto x^2$ shows.

*under the additional assumption that $\lim_\infty f$ exists (name it $\ell\in\mathbb{R}$), then this becomes a standard exercise (most likely asked several times on this website). The idea is to use


*

*the fact that for $x,y$ "big", then $\lvert f(x) - f(y) \rvert \leq \lvert f(x)-\ell\rvert +  \lvert \ell - f(y)\rvert$ and both terms of the RHS are very small since $f\to_\infty \ell$;

*and for $x,y$ "small", well they are both in a compact interval of the form $[a,A]$, and then anyway $f$ is uniformly continuous on this compact by the Heine—Cantor theorem.


More precisely, here is the outline:


*

*for every $\varepsilon>0$ you have a value $A_\varepsilon$ such that $f(x)\in[\ell-\frac{\varepsilon}{2},\ell+\frac{\varepsilon}{2}]$, and therefore by the above $f(x)$ and $f(y)$ satisfy $\lvert f(x) - f(y) \rvert \leq \varepsilon$ whenever $x,y\in I=[A_\varepsilon,\infty)$.

*But then, $[a,A_\varepsilon]=J$ is a compact, so $f$ is uniformly continuous on $J$, and there exists $\delta > 0$ such that for any $x,y\in J$ such that $\lvert x-y\rvert < \delta$ we have $\lvert f(x) - f(y) \rvert \leq \varepsilon$ as well.

*The last case to deal with is when $x\in I,y\in J$: then, use continuity at $A_\varepsilon$.
Combining the three suitably, you'll prove (as $\varepsilon$ was arbitrary) that $f$ uniformly continuous on $[a,\infty)$. 
