Show that if $f:[a, \infty) \rightarrow \mathbb{R}$ is continuous and $\lim \limits_{x \to \infty} f(x) < \infty$, then $f$ is uniformly continuous at $[a, \infty)$.
My attempt:
Let $c \in A=[a, \infty)$. Since $f$ is continuous at $A$, it means $f$ is continuous for all $c \in A$.
Since $f$ is continuous at $c\in A$, it means $\forall \varepsilon > 0 \,\exists \delta > 0$ such that $\forall x$ if $|x-c| < \delta$ then $|f(x)-f(c)|<\varepsilon$.
Let $\lim \limits_{x \to \infty} f(x) = L < \infty$, it means $\forall \varepsilon > 0 \,\exists \delta > 0$ such that $\forall x$ if $0< |x-c| < \delta$ then $|f(x)-L|<\varepsilon$.
So, I don't know what should I do next to show $f$ is uniformly continuous at $[a, \infty)$. Please help, thank you!
Let $\epsilon>0$. Need to find $\delta>0$ such that for all $x,y\in [a,\infty)$, if $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon$. Say $\lim_{x\to\infty} f(x)=L$. Then there is some $x_0\in [a,\infty)$ such that for all $x>x_0$ we have $|f(x)-L|<\epsilon/2$. Now assume $x,y\in (x_0,\infty)$. Then we have $|f(x)-f(y)|=|f(x)-L+L-f(y)|\le |f(x)-L|+|f(y)-L|<\epsilon/2+\epsilon/2=\epsilon$. Finish the proof by showing $f$ is uniformly continuous on $[a,x_0]$.
Hints:
$[a,x_0]$ is a compact set.
To make $|f(x)-f(y)|<\epsilon$ when $x,y>x_0$ we didn't need a specific $\delta$.