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Given the following function: $$x\sin\bigg(\frac{1}{x}\bigg)$$

How do I show that this is uniformly continuous on $(0, \infty)$?

The reason why I am having trouble is that I am use to finding pathologies in functions which causes them to not have a certain property.

Furthermore, is there a more "analysis" like way of doing this.

(1) $x\sin(\frac{1}{x})=\frac{\sin(\frac{1}{x})}{\frac{1}{x}}$

(2) L'Hospital's Rule: $\sin(\frac{1}{x})-\frac{\cos(\frac{1}{x})}{x}$. Taking all appropriate limits I get $1$.

This is about where I'm stuck.

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1 Answer 1

up vote 7 down vote accepted

As $-x \le x \sin (x^{-1}) \le x$, your function has a limit at 0, namely $0$, so we may extend it to a continuous function $f\colon [0,\infty) \to \mathbb R$, which is uniformly continuous on $[0,2]$ (as this is a compact interval) and on $[1,\infty)$ (as $f$ has bounded derivative \[ f'(x) = \sin\left(\frac 1x\right) - \frac{\cos\left(\frac 1x\right)}{x} \] here and is therefore Lipschitz). Hence $f$ is uniformly continuous on $[0,\infty)$, hence on $(0,\infty)$.

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