Uniform continuity of $x \sin{\frac{1}{x}}$

Question

$f(x) = x \sin{\frac{1}{x}}$ for $x > 0$, $0$ otherwise. Prove $f$ is uniformly continuous. I know similar questions have appeard asking for uniform continuity on $(0, 1)$, but here is uniform continuity on $\mathbb{R}$ and such problems are sensitive to the domain. I am wondering how to prove it formally.

Answer 1

As an easy exercise, you can prove that if $f:[0,\infty )\to \mathbb R$ is continuous and $$\lim_{x\to \infty }f(x)=f(0),$$ then $f$ is uniformly continuous.

Hint

Let $\varepsilon >0$.

1) Since $\lim_{x\to \infty }f(x)=f(0)$, there is $M>0$ s.t. $x\geq M\implies |f(x)-f(0)|<\frac{\varepsilon }{2}$.

2) If $x,y\geq M$, it's straightforward that $|f(x)-f(y)|\leq \varepsilon $ using the previous point.

3) If $0\leq x,y\leq M$, it should be a basic theorem.

4) Try to see what happen when $x\leq M\leq y$ and $y\leq M\leq x$,

5) Conclude.