Prove that the function \(f(x)=\frac{\sin(x^3)}{x}\) is uniformly continuous and bounded.

TIFR 2019 GS Part A, Problem 9

Analysis

Hard

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We have the function in the interval \((0,\infty)\) \(f(x)=\frac{\sin(x^3)}{x}$\) Can you prove that the function is uniformly continuous?

The function is clearly continuous(why?).

Any bounded, continuous function where as is uniformly continuous. The derivative if it exists does not have to be bounded.

Note that \(\sin(x^3)/x = x^2 \sin(x^3)/x^3 \to 0\cdot 1 = 0\) as \(x \to 0\). This is also a great example of a uniformly continuous function with an unbounded derivative

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