Let $f : [a, +\infty) \rightarrow \mathbb{R}$ be continuous, such that $\lim_{x \rightarrow +\infty} f(x) = l \in \mathbb{R}$. Show that f is uniformly continuous.
I want to split the domain - one in which we have continuity, and the other in which the values of f are approaching $l$. This means there are theree possibilities for our $x$ and $y$ using the uniformly continuous definition: x and y are in the first part, x and y are in the second, x is in the first and y in the second.
How would I do this proof?
It suffices you do the following. Given $\epsilon>0$, take $N$ large so that, for $x>N$,
$$|f(x)- l|<\varepsilon/2$$
It follows that if $x,y>N$, then $|f(x)-f(y)|<\varepsilon$. Now consider the interval $[a,N+1]$, which is compact. Here $f$ is uniformly continuous, so there is some $\delta >0$ such that if $|x-y|<\delta$ then $|f(x)-f(y)|<\varepsilon$. We can assume that $\delta<1$.
Consider now arbitrary $x,y\geqslant a$ that are at most $\delta$ units apart. If both $x,y\leqslant N+1$, then you're done. Our choice of $\delta$ is such that if $x> N+1$ then also $y>N$, and you're also done.