# If a function $f$ is continuous in $[a,∞)$ and finite $\lim_{x→+∞}⁡f(x)$ exists, then it's uniformly continuous in $[a,+∞)$.

Prove that if $f$ is defined and continuous in $[a,+∞)$ and if there exists a finite limit $\lim_{x→+∞}⁡f(x)$, then $f$ is uniformly continuous in $[a,+∞)$

I know that since there exists a finite limit $\lim_{x→+∞}⁡f(x)$, so $f(x)$ is convergent. How can I use this to prove it's uniformly continuous?

• This question was asked recently. Oct 26, 2013 at 14:13
• @njguliyev the other question is to prove that $f$ is bounded, not continuous Oct 26, 2013 at 14:16
• Sorry. You are right. Oct 26, 2013 at 14:22
• Feb 16, 2014 at 12:01

## 2 Answers

The idea is to reduce the proof to the compact case.

Let $L= \lim_{x\to\infty} f(x)$. Choose $\epsilon>0$, and $M>0$ such that if $x>M$, then $|f(x)-L| < \frac{1}{2}\epsilon$.

$f$ is continuous, hence uniformly continuous on $[a,M+1]$. Choose $0<\delta<1$ so that if $x,y \in [a,M+1]$ and $|x-y| < \delta$, then $|f(x)-f(y)| < \epsilon$.

Now choose $x,y \in [a,\infty)$ such that $|x-y| < \delta$. Then either $x,y \in [a,M+1]$ or $x,y \in (M, \infty)$. In the first case we have $|f(x)-f(y)| < \epsilon$, in the second case, we have $|f(x)-f(y)| \le |f(x)-L|+|f(y)-L|< \frac{1}{2}\epsilon + \frac{1}{2}\epsilon = \epsilon$.

• Hi @copper.hat why did you choose $0<\delta<1$, and did you meant $[a,M]$ instead of $[a,M+1]$?
– user864806
Dec 10, 2021 at 18:36
• @Bestmat When I select $x,y \in [a,\infty)$ I want to either have both in $[a,M+1]$ or both in $(M,\infty)$ so that the earlier bounds can be used. Dec 10, 2021 at 18:40
• Yes, but why $\delta<1$ @copper.hat is it because we want the two points to be in the same interval?
– user864806
Dec 10, 2021 at 18:48
• @Bestmat Yes. ${}{}$ Dec 10, 2021 at 19:10

Let $L=\lim_{x\to\infty}f(x)$. Given $\epsilon>0$, there is $b\in[a,\infty)$ such that $|f(x)-L|<\frac12\epsilon$ for all $x>b$. Hence $|f(x_1)-f(x_2)|<\epsilon$ if $x_1,x_2\in[b,\infty)$. Since $[a,b+1]$ is compact, $f$ is uniformly continuous on $[a,b]$, hence there exists $\delta_b>0$ such that $|x_1-x_2|<\delta_b$ with $x_1,x_2\in[a,b+1]$ implies $|f(x_1)-f(x_2)|<\epsilon$. Then for $x_1,x_2\in[a,\infty)$ with $|x_1-x_2|<\delta:=\min\{\delta_b,1\}$ we have that $x_1,x_2\in[a,b+1]$ or $x_1,x_2\in[b,\infty)$, hence by one of the results just shown $|f(x_1)-f(x_2)|<\epsilon$.