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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.

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

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    $\begingroup$ @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 $\endgroup$ – Federico Dec 10 '18 at 19:08
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    $\begingroup$ @Federico You are right. before answering, i had to think twice. $\endgroup$ – hamam_Abdallah Dec 10 '18 at 19:10
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    $\begingroup$ @hamam_Abdallah Hey, cheer up! ;) $\endgroup$ – Federico Dec 10 '18 at 19:12

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