Question: Let $f\in C([0,\infty))$. And $\forall \ h\in\mathbb{R}$,
$$
\lim_{x\to\infty}|f(x+h)-f(x)|=0.
$$
Show that $f$ is uniformly continuous on $[0,\infty)$.
I have some idea about this question, but I can not solve it.
To show $f$ is uniformly continuous on $[0,\infty)$, we claim that: $$ \forall \ ε>0, \ \exists\ M_{ε}>0, \ \exists \ r_{ε}>0, \ \forall \ x>M_{ε}, \ \forall \ h\in[0,r_{ε}], \ |f(x+h)-f(x)|<ε. $$ If our claim is true, then $f\in C([0,M_{ε}+r_{ε}])$ gives a $\eta_{ε}>0$ such that $\forall \ x,y\in [0,M_{ε}+r_{ε}]$ with $|x-y|<\eta_{ε}$, $|f(x)-f(y)|<ε$. Set $\delta=\min\{\frac{1}{2}r_{ε},\eta_{ε}\}$, then $\forall \ x,y\in [0,\infty)$ with $|x-y|<\delta$, $|f(x)-f(y)|<ε$. Hence we get that $f$ is uniformly continuous on $[0,\infty)$. So we just need to show our claim.
Suppose our claim is not true: $$ \exists\ ε_{0}>0, \ \forall \ r>0, \exists \ x_{n}>n, \ \exists \ \{h_{n}\}\subset[0,r] \mbox{ such that }|f(x_{n}+h_{n})-f(x_{n})|\geqslant ε_{0}. $$ Then I want to get $h\in\mathbb{R}$ and $\{x_{n_{k}}\}\subset\{x_{n}\}$ such that $\lim\limits_{k\to\infty}|f(x_{n_{k}}+h)-f(x_{n_{k}})|\not=0$. But how to prove this? I have not idea about this...
