# Let $f:(0,\infty)\to\mathbb{R}$ 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.

• No, option (a) is not correct; the numerator vanishes at $0$ as well, and does so faster than the denominator. – user296602 Dec 10 '18 at 18:04
• yes , you are right then this becomes unbounded in $(0,\infty)$ – sejy Dec 10 '18 at 18:06
• @sejy No it doesn't. Think about what T. Bongers said - how could $\sin(x^3)/x$ be unbounded in $(0,+\infty)$. Can you show me an $x$ such that $\sin(x^3)/x=2$? – Jam Dec 10 '18 at 18:12
• $-\frac{1}{x} < \frac{\sin{x^3}}{x} < \frac{1}{x}$ on $[1,\infty]$. All that's left is that pesky $(0,1)$ region. – David Diaz Dec 10 '18 at 18:22
• A lot of confusion here. This function is very well behaved, in fact $\int_0^\infty \frac{\sin x^3}{x} \, dx = \frac{\pi}{6}$ – RRL Dec 10 '18 at 18:53

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.

• @hamam_Abdallah I mean, it is trivial to see that near $0$ you have $\sin(x^3)/x\sim x^3/x=x^2$, so it is definitely bounded there – Federico Dec 10 '18 at 19:08
• @Federico You are right. before answering, i had to think twice. – hamam_Abdallah Dec 10 '18 at 19:10
• @hamam_Abdallah Hey, cheer up! ;) – Federico Dec 10 '18 at 19:12