What are different possible ways of proving that $\frac{\sin x}{x}$ is uniformly continuous?

Question

My main preoccupation is to find different Method to prove that the function

$$ f :x\mapsto \frac{\sin x}{x}$$ is uniformly continuous on $\mathbb R$. Particularly I am not able to prove it direct using the $\epsilon-\delta $ definition . Here What is I found so far. Since $f$ is continuous and $f(x)\to 0$ as $|x|\to \infty $ we conclude that $f$ is uniformly continuous using this result But I believe that there are others ways to overcome this this issue.

Edit: Note that I am asking different possible way to to prove the uniform continuity of $f.$

It is not defined over $\mathbb{R}$ and therefore cannot be uniformly continuous over it... — John Do, Oct 06 '17 at 12:13
Differentiate, note that the derivative is bounded, and conclude that $f$ is even Lipschitz-continuous. — Daniel Fischer, Oct 06 '17 at 12:21
@JessePFrancis My Question completely different from what you are suggesting . how do you find duplicate? Please red carefully — Guy Fsone, Oct 06 '17 at 13:34
This question is wrongly marked as duplicate because the accepted answer to the other thread is that a continuous function on a compact space is uniformly continuous. This answer is irrelevant here as ${\mathbb R}$ is not compact. — Gribouillis, Oct 06 '17 at 14:33
They should be more careful. Observe that the message above says 'if those answers do not fully address your question, please ask a new question'. In this case, the other answer does not at all address the question, but there is no point in asking a new question, because it is not a duplicate! — Gribouillis, Oct 06 '17 at 14:48
Perhaps a better choice for duplicate would be Determine whether $f(x)=\sin x/x$ is uniformly continuous in $\Bbb R$ — Joffan, Oct 06 '17 at 15:26
@Joffan Thanks. It's indeed a duplicate of that one. I've changed the target to it. — Daniel Fischer, Oct 06 '17 at 16:04
My bad. I didn't notice the (0,1) bit in the question I tagged as duplicate. — Jesse P Francis, Oct 07 '17 at 01:07

score 11 · Accepted Answer · edited Oct 06 '17 at 13:06

11

One has $$f(x) = \frac{\sin(x)}{x} = \int_0^1 \cos(x t)\,d t$$ hence $$ \left|f^\prime(x)\right| = \left |\int_0^1 t \sin(x t)\,dt\right | \le\int_0^1t\,d t \le \frac{1}{2} $$ hence $$\forall x, y\quad |f(x) - f(y)| \le \frac{1}{2}|x-y|$$

edited Oct 06 '17 at 13:06

gen-ℤ ready to perish

6,732

answered Oct 06 '17 at 12:25

Gribouillis

16,826

This is a more elegant proof – Guy Fsone Oct 06 '17 at 13:29

score 2 · Answer 2 · answered Oct 06 '17 at 12:36

For an intuitive approach: Note that we always have the mean value inequality $$ \bigg |\frac{\sin x}{x}-\frac{\sin y}{y}\bigg|\leq\sup_{c\in\mathbb{R}}\bigg|\frac{c\cos c-\sin c}{c^2}\bigg||x-y| $$ but the right hand side is clearly finite. The only interesting part is near $c=0$, where we might be concerned we have a blow up. Here, Taylor expansions show us derivative is roughly 1.

What are different possible ways of proving that $\frac{\sin x}{x}$ is uniformly continuous?

2 Answers2