Thursday, 21 March 2019

real analysis - Let f:(0,infty)tomathbbR be defined by f(x)=fracsin(x3)x. Then f is not bounded and not uniformly continuous.



Let f:(0,)R be defined by f(x)=sin(x3)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 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.


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