# Understand the problem

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

##### Topic

Analysis

##### Difficulty Level

Hard

##### Suggested Book

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

# Start with hints

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 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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