Analyse Equicontinuity (uniform continuity) of a function $ \frac{\sin(x)}{x},\; x > 0 $

Question

How can I analyse equicontinuity of the function $$ \frac{\sin(x)}{x},\; x > 0? $$

When I draw the graph it is clear that this function is equicontinous, but I can not prove it analytically. Tried to prove using definition. I am stuck. Any hints or better ways to do this kind of problems?

Answer 1

This is about uniform continuity. "Equicontinuity" refers to families of functions: All members of the family should behave equally nicely.

Here we are given the function $${\rm sinc}(x):=\left\{\eqalign{{\sin x\over x}\quad&(x\ne0)\cr 1\qquad&(x=0)\ .\cr}\right.$$ Note that we can write this function in the form $${\rm sinc}(x)=\int_0^1\cos(t x)\>dt\qquad(-\infty<x<\infty)\ ,$$ so that $$\bigl|{\rm sinc}(y)-{\rm sinc}(x)\bigr|\leq\int_0^1\bigl|\cos(ty)-\cos (tx)\bigr|\>dt\leq\int_0^1\bigl|ty-tx\bigr|dt={1\over2}|y-x|\ .$$ This shows that the ${\rm sinc}$ function is Lipschitz continuous, hence uniformly continuous, over all of ${\mathbb R}$.

Answer 2

If $f$ is continuous on $\mathbb R$ and $\lim_{|x|\to \infty}f(x)=0,$ then $f$ is uniformly continuous on $\mathbb R.$ Clearly $f(x)=(\sin x)/x$ satisfies these criteria.