proved that$f (x)$ is uniformly continuous on $[0,+\infty)$ if and only if $\lim\limits_{x\to+\infty}\frac{f(x)}{x}$ exists. Let $f(x)$ be second-order differentiable on $[0,+\infty)$,$f''(x)\geqslant\sin x(\forall x\in[0,+\infty))$.Prove that $f (x)$ is uniformly continuous on $[0,+\infty)$ if and only if $\lim\limits_{x\to+\infty}\frac{f(x)}{x}$ exists.This is our monthly exam question,I try to expand it with Taylor's theorem:
$$f(x+h)=f(x)+f'(x)h+\frac{f''(\xi)}{2}h^2\geqslant f(x)+f'(x)h +\frac{\sin x}{2}h^2.$$
But then I don't know how to deal with it, I also know that if $f (+\infty)$ exists, then $f (x)$ is uniformly continuous, but it doesn't seem to help the problem.

Some comments:
(1) We cannot drop the continuity from the right at $x=0$. Here is a counterexample: the function $f(x):= \frac{1}{x}-\sin{x}$ is not uniformly continuous on $(0,\infty)$. We have $f^{\prime\prime}(x)=\frac{2}{x^3}+\sin{x}>\sin{x}$ for every $x\in (0,\infty)$. And obviously $\lim_{x\rightarrow \infty}\frac{f(x)}{x}=0$.
(2)   It is also enough to prove that $f$ is uniformly continuous on $[1,\infty[$ since it is, by assumption, continuous on any closed interval
$[0,N]$ with $N>0$, and consequently uniformly continuous thereat.
(3)  We know that, since $\lim_{x\rightarrow +\infty}\frac{f(x)}{x}$ exists,
$$|f(x)|\leq C |x|$$
for large $x$, meaning the function can grow at most linearly.
But the statement does not apply to functions of the form $f(x)=x^\alpha$ for any $\alpha\in (0,1]$ because the restriction $f^{\prime\prime}(x)\geq \sin{x}$ is not satisfied by such functions.
(4) I suggest the following simplification of the problem:
We can write any smooth $f:[0,\infty)\rightarrow \mathbb{R}$ as
$f(x)=g(x)-\sin{x}$ where $g:[0,\infty)\rightarrow \mathbb{R}$
is smooth. Now, the peculiar condition $f^{\prime\prime}(x)\geq \sin{x}$
translates into $g^{\prime\prime}(x)\geq 0$. We also have $\lim_{x\rightarrow +\infty}\frac{g(x)}{x}=\lim_{x\rightarrow +\infty}\frac{f(x)}{x}$ exists. Since $x\mapsto \sin{x}$ is uniformly continuous on $\mathbb{R}$, we only need to show that $g$ is uniformly continuous on $[N,\infty)$ with $N$ large.
