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

Continuous function in the interval $$(0,\infty)$$ $$f(x)=\frac{\sin(x^3)}{x}$$. To prove that the function is uniformly continuous. The function is clearly continuous. Now $$|f(x)-f(y)|=|\frac{\sin(x^3)}{x}-\frac{\sin(y^3)}{y}|\leq |\frac{1}{x}|+|\frac{1}{y}|$$. But I don't think whether this will work.

I was trying in the other way, using Lagrange Mean Value theorem so that we can apply any Lipschitz condition or not!! but $$f'(x)=3x^2\frac{\cos(x^3)}x-\frac{\sin(x^3)}{x^2}$$

Any hint...

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 $$x \to 0$$.
• I think in your proposition you also need $f(x)\to 0$ as $x\to 0$. Because the idea of the proof is to use that in a compact set $[\epsilon, N]$ your function is uniformly continuous and in the intervals $(0,\epsilon)$ and $(N,\infty)$ you can control the difference $|f(x)-f(y)|$since $f$ is small here. – Julian Mejia May 13 '19 at 5:27
• Basically finite limits at both ends is what is needed, not necessarily $0$. Having overlooked that point in writing a hint gives the downvoter an opportunity. – RRL May 13 '19 at 5:34