uniform continuity only using theorems

Question

$f : [0,\infty) \to \mathbb{R}$ be continuous on $[0,\infty)$ and $\lim\limits_{x \to \infty} f(x)=0$. Prove that f is uniformly continuous on $[0,\infty)$ . Can this be proved by using the theorems only, not using the definition directly?

I approach to the answer like this : since $\lim\limits_{x \to \infty} f(x)=0$ holds, so $f(x)$ is bounded everywhere in $[0,\infty)$. Now $[0,1],[1,2],[2,3].....$ are closed and bounded, so $f(x)$ is uniformly continuous on each of $[0,1],[1,2],[2,3].....$ now $[0,\infty)=[0,1]\cup[1,2]\cup[2,3].....$ so $f(x)$ is uniformly continuous on $[0,\infty)$ . I approach this way only because $f(x)$ is bounded. the But I don't understand if I am right or not. Is this approach correct ?

Answer 1

Maybe you want to make use of the theorem that each continuous function on a compact subset of $\mathbb R$ is uniformly continuous. Then choose $n>0$ so big that for $x>n$ we have $|f(x)|<\varepsilon/2$. For $[0,n+1]$ use the theorem. QED.

Answer 2

Say a function $g$ has compact support if there exists $A$ so that $g(x)=0$ for all $x$ with $|x|>A$.

Theorem: A continuous function with compact support is uniformly continuous.

Theorem: If $f_n$ is uniformly continuous and $f_n\to f$ uniformly then $f$ is uniformly continuous.

Theorem: If $f$ satisfies the conditions you give then there exist continuous functions $f_n$ with compact support such that $f_n\to f$ uniformly.