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

Question

I am asked whether the function $f(x)=x\sin(\frac{1}{x})$ is uniformly continuous on $\mathbb{R}\setminus\{0\}$

My first intuition was to compute its derivative: $f'(x)=\sin(\frac{1}{x})-\frac{1}{x}\cos(\frac{1}{x})$

I found out that this function is not bounded when approaching $0$. So $f$ is not Lipschitz on $\mathbb{R}\setminus\{0\}$

I then picked sequences $x_{n}=\frac{1}{2\pi n}$ and $y_{n}=-\frac{1}{2\pi n}$ and computed $|f(x_{n})-f(y_{n})|=4\pi n\rightarrow+\infty$

So I am convinced that $f(x)$ is not uniformly continuous on $\mathbb{R}\setminus\{0\}$

Is my approach correct?

Answer 1

No. The function $f$ is uniformly continuous. Hint: near $0$ because it is continuous in, say, $[-1,1]$ and far from $0$ because then $f'$ is bounded.