$f :

This question already has an answer here:

  • How does the existence of a limit imply that a function is uniformly continuous

    5 answers

Solutions Collecting From Web of "$f :"

Suppose $\lim_{x\to \infty} f(x) = L$, then for any $\epsilon >0$, there is $N \in \mathbb{R}$ such that
$$
x \geq N \Rightarrow |f(x) – L| < \epsilon/3
$$
Hence, for any $x,y > N$
$$
|f(x) – f(y)| < \epsilon/3
$$
Now, $f$ is uniformly continuous on $[a,N]$, so there is a $\delta > 0$ such that
$$
|x-y|<\delta, \text{ and } x,y \in [a,N]\Rightarrow |f(x) – f(y)| <\epsilon/3
$$
Hence, for any $x,y \in [a,\infty)$, if $x, y \in [a,N]$ or $x,y \geq N$, then
$$
|x-y|<\delta \Rightarrow |f(x) – f(y)| < \epsilon \qquad\text{(1)}
$$
Furthermore, if $x < N < y$, then $|x-y|<\delta$ implies that
$$
|x-N| < \delta, \text{ and } |f(y) – L| < \epsilon/3, |f(N) – L|<\epsilon/3
$$
hence
$$
|f(x)-f(y)| \leq |f(x) – f(N)| + |f(N) – L| + |L – f(y)| < \epsilon
$$
Hence, (1) holds for all $x,y\in [a,\infty)$