# Uniform continuity of $x\sin(\frac{1}{x})$ [duplicate]

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?

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.
• Thank you for your answer! I actually did a stupid computation mistake (plugged in $f'(x)$ values instead of $f(x)$ when using sequences). I like the strategy you propose. So, I can easily show that $f$ is continuous at $0$ by the definition. Then, I can go on to show that it is continuous on $[-1,1]$ by virtue of being the product of continuous functions. Finally, use your $f'$ is bounded argument for $|x|\ge1$, which means it is Lipshitz and therefore uniformly continuous. Since $[-1,1]$ and $[1,+\infty]$ have {$1$} as intersection (and the same for -$infty$), then it is unif on $\mathbb{R}$ May 12 '18 at 23:22