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 $f.$

Answer 1

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 $$\forall x, y\quad |f(x) - f(y)| \le \frac{1}{2}|x-y|$$

Answer 2

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.