Is $f(x)=\sin(x^3)/x$ uniformly continuous?

Question

Let $f : (0, \infty) \to \mathbb R$ be defined by $f(x)=\sin(x^3)/x$. Then, can we say that $f$ is uniformly continuous on $(0, \infty)$?

Since $\lim\limits_{x \to 0}f(x)$ exists, we can ensure that uniformly continuous in any $[0,a]$, $a>0$. However, what is the nature of $f(x)$ at larger $x$?

Answer 1

Given $n$, assume that $x,\ y>n$ Then $|f(x) -f(y)| <\frac{1}{x} + \frac{1}{y} < \frac{2}{n}$

When $x,\ y<\frac{1}{n}$, then $|f(x)-f(y)|<\frac{2}{n^2}<\frac{1}{n}$

So now consider $f$ on $[1/n,n]$. Hence we have uniform continuity

Answer 2

The numerator is bounded so $f(x) \to 0$ as $x \to \infty$. This gives uniform continuity.