Uniform continuity of $f(x)=\frac{1}{x}\cos(2\pi x^3)$ 
Let $f:(0,+\infty)\rightarrow\mathbf{R}$ be defined by $$f(x)=\frac{1}{x}\cdot\cos(2\pi x^3)$$
  Prove that $f$ is uniformly continuous on $[1,+\infty)$.

In previous questions, I proved that $f$ is continuous and that $f$ is uniformly continuous on $[1,R]$ for $R>1$.
The derivative is not bounded, so we can't use that theorem. I tried proving it directly using the definition, but that won't work. Could someone provide any help?
 A: The following more general fact holds.

If $f: [1,\infty) \rightarrow \mathbb{R}$ is a continuous function such that $\lim_{x \to \infty} f(x) = 0$ then  $f$ is uniformly continuous on $[1,\infty)$. 

Let $\epsilon>0$. The limit $\lim_{x \to \infty} f(x) = 0$ implies that there is $R\geq 1$ such that $|f(x)|<\epsilon/4$ for $x\in [R,+\infty)$.
Moreover, you already know that $f$ is uniformly continuous in $[1,R]$, therefore there is $\delta>0$ such that $|f(x)-f(y)|<\epsilon/2$ for $|x-y|<\delta$ and $x,y\in [1,R]$. 
Now we show that $f$ is uniformly continuous on $[1,\infty)$: 
for $x,y\in [1,\infty)$, such that $|x-y|<\delta$, then $|f(x)-f(y)|<\epsilon$.
Assume $1\leq x\leq y$ and consider three cases. 
1) If $x,y\leq R$ then $|f(x)-f(y)|<\frac{\epsilon}{2}<\epsilon.$
2) If $x\leq R\leq y$ then $|x-R|<\delta$ and 
\begin{align*}
|f(x)-f(y)|&\leq |f(x)-f(R)|+|f(R)-f(y)| \\
&\leq |f(x)-f(R)|+|f(R)|+|f(y)|< \frac{\epsilon}{2}+\frac{\epsilon}{4}+\frac{\epsilon}{4}=\epsilon.
\end{align*}
3) If $R\leq x\leq y$ then $|f(x)-f(y)|\leq |f(x)|+|f(y)|<\frac{\epsilon}{4}+\frac{\epsilon}{4}<\epsilon.$
