[Math] Determine whether $f(x)={\sin x \over x}$ is uniformly continuous in $\mathbb R$

Question

Determine whether the function$$f(x)={\sin x \over x}$$is uniformly continuous in $\mathbb{R}$.

I am using the definition that for $\epsilon>0$ there exists $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$.

Answer 1

One has $$f(x):={\sin x\over x}=\int_0^1 \cos(\tau x)\ d\tau\ .$$ It follows that $$|f(x)-f(y)|\leq \int_0^1|\cos(\tau x)-\cos(\tau y)|\ d\tau\leq \int_0^1\tau |x-y|\ d\tau={1\over2}|x-y|\ .$$ This proves that $f$ is even Lipschitz-continuous.