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

Source of the problem
TIFR 2019 GS Part A, Problem 9
Analysis
Hard

Introduction to Real Analysis by Donald R. Sherbert and Robert G. Bartle

Do you really need a hint? Try it first!

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 $f:(0,\infty) \to \mathbb{R}$ where $f(x) \to 0$ as $x \to 0,\infty$ 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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