$f:[a,\infty) \longrightarrow \mathbb{R}$ continuous and $\lim_{x\to \infty}{f(x)}=c\in \mathbb{R}$ so $f$ is uniformly continuos in $[a,\infty)$

Question

let be $ \ f:[a,\infty) \longrightarrow \mathbb{R} \ $. Such that f is continuous in $[a,\infty)$ and: $$\lim_{x\to \infty}{f(x)}=c\in \mathbb{R}.$$

then $f$ is uniformly continuous in $[a,\infty)$.

I would like to know how to prove it by the defininiton of uniform continuity or by sequences characterization of uniform continuity.

Than you so much.

Answer 1

Let $\epsilon >0$. Since $\lim_{x \to \infty} f(x)= c$, then there is $A>0$ such that

$$|f(x) -c| < {\epsilon \over 4},\,\,\, \forall x >A.$$

Choose $M>A>0$. Then for every $x,y \in [a,\infty)$ we have

If $x,y \in [a, M]$, then the compactness of the closed interval we have the fact that continuity implies uniformly continuity inside the closed interval.

If $x,y > M$ then

$$|f(x) - f(y)| \leq |f(x) -c| + |c -f(y)| < {\epsilon \over 2} < \epsilon.$$

Finally, when $x \in [a,M]$ and $y >M$, we have

$$|f(x) -f(y)| \leq |f(x)-f(M)| + |f(M)- c| + |c -f(y)| <\epsilon.$$

Above we assumed $|f(x)-f(M)| < \frac \epsilon 2$ because the uniform continuity inside the closed interval $[a,M]$.