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.

proof:
We think $f^{''}(x)=sin(x)+g(x)$. Cuase $f^{''}(x)>sin(x)$, $g(x)>0$ on $[0,\infty)$.
So we can get $f(x)=-sin(x)+(f^{'}(0)+1)x+f(0)+\int_{0}^{x}\,dy \int_{0}^{y} g(t)\, dt$
It's easy to konw $f(x)$ being uniformly continuous on $[0,\infty)$ is equivalent to $f^{'}(x)$ being bounded, which is equivalent to $\int_{0}^{\infty} g(x)\,dx$ is convergent, under the conditions given in the problem.
It's easy to find that $\lim_{x \to \infty} \frac{f(x)}{x}=\lim_{x \to \infty} \frac{-sin(x)+(f^{'}(0)+1)x+f(0)+\int_{0}^{x}\,dy \int_{0}^{y} g(t)\, dt}{x}=f^{'}(0)+1+\lim_{x \to \infty}\frac{\int_{0}^{x}\,dy \int_{0}^{y} g(t)\, dt}{x}=f^{'}(0)+1+\int_{0}^{\infty} g(t)\,dt$
This shows that $\lim_{x \to \infty} \frac{f(x)}{x}$ being exist is equivalent to $\int_{0}^{\infty} g(x)\,dx$ being convergent, which means that $\lim_{x \to \infty} \frac{f(x)}{x}$ being exist is equivalent to $f(x)$ being uniformly continuous on $[0,\infty)$.
