Problem
Suppose
$f:\Bbb R^+\to\Bbb R$ satisfies $\forall\epsilon>0,\exists E>0,\forall x_0>E,\exists\delta>0,\forall x(\left|x-x_0\right|<\delta): \left|f(x)-f(x_0)\right|<\epsilon\tag1$
Can we conclude that
there's some continuous function $g:\Bbb R^+\to\Bbb R$ such that $\lim_{x\to+\infty}(f(x)-g(x))=0\tag2$
Re-describe
Let $\omega_f(x_0)=\limsup_{x\to x_0}\left|f(x)-f(x_0)\right|$, we can re-describe the first condition (1) as this: $\lim_{x_0\to+\infty}\omega_f(x_0)=0\tag3$
Motivation
In fact, I'm discovering the sufficient and necessary condition of (2). It's easier to show that (2) implies (3), i.e. (1), because $\left|f(x)-f(x_0)\right|\le\left|f(x)-g(x)\right|+\left|g(x)-g(x_0)\right|+\left|g(x_0)-f(x_0)\right|$ Take $\lim_{x_0\to+\infty}\limsup_{x\to x_0}$ for both sides, we'll get the result.