real analysis - Let $f:(0,infty)tomathbb{R}$ be defined by $ f(x)=frac{sin(x^{3})}{x}$. Then f is not bounded and not uniformly continuous.

Let $f:(0,\infty)\to\mathbb{R}$ be defined by $ f(x)=\frac{\sin(x^{3})}{x}$. Then which of the following is correct:

a)f is not bounded and not uniformly continuous

b)f is bounded and not uniformly continuous

c)f is not bounded and uniformly continuous

d)f is bounded and uniformly continuous

I think option a is correct $\because \sin{x}$ is bounded between $-1$ and $1$ and $\frac{1}{x}$ approches $\infty$ in neighborhood of zero.

This question was asked in TIFR 2019.

Answer

The function is bounded and uniformly continuous on $(0,\infty)$.

Clearly, $f$ is continuous and, hence, uniformly continuous on any compact interval $[a,b]$ with $a > 0$.

On the interval $(0,a]$ we have $\displaystyle f(x) = \frac{\sin x^3}{x} = x^2\frac{\sin x^3}{x^3} \to 0\cdot 1 = 0 $ as $x \to 0$ and
$f$ is extendible as a continuous function to the compact interval $[0,a]$, and, hence, uniformly continuous there.

On $[b, \infty)$, $f$ is uniformly continuous as well since $\displaystyle |f(x)| = \frac{|\sin x^3|}{x} \leqslant \frac{1}{x} \to 0 $ as $x \to \infty$.

A continuous function that approaches a finite limit as $x \to \infty$ must be uniformly continuous -- proved many times on this site -- for example here. This is also an interesting example of a function with an unbounded derivative that is uniformly continuous.