uniformly continuous functions on infinite open interval $(0, \infty)$

Question

Is $\sin x/x$ uniformly continuous on $(0, \infty)$? Can any one help me proving this? $d/dx(\sin x/x)$ is not bounded on $(0, \infty)$. How to proceed then?

Answer 1

The derivative is bounded. We have $$\frac{d}{dx}\frac{\sin x}{x}=\frac{x\cos x-\sin x}{x^2}.$$ Now, use that for $x> 0$ $$\sin x\leq x-\frac{x^2}{2},\quad\text{and}\quad\cos x\leq 1.$$ This gives $$\frac{d}{dx}\frac{\sin x}{x}\leq\frac{x-(x-x^2/2)}{x^2}=\frac{1}{2}.$$ Now it is easy to show that $\frac{d}{dx}\frac{\sin x}{x}$ on $(0,\infty)$ extends continuously to $[0,\infty)$, so $1/2$ is a global bound on $[0,\infty)$. Can you take it from there?