Suppose that $f : [a, ∞) → R$ is a continuous function. If $\lim\limits_{x→∞} f (x) = L$, prove that $f$ is uniformly continuous on $[a, ∞)$.
My attempt at the proof:
Well since I have to use both facts, I think my proof needs to be divided into two parts:
1- I have to prove that $f$ is uniformly continuous on $(N,∞)$ where $N>a$ (by using the limit definition somehow)
2- Using compactness, I can easily show that $f$ is uniformly continuous on $[a,N]$
I am having trouble with $(1)$ because I can't find an $\delta>0$ that works for all $\epsilon>0$
Just need a hint in the right direction, thank you!!
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 $x,y\in [a,\infty)$