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Let $f(x):\mathbb{R} \rightarrow \mathbb{R}$ be a continuous functionsuch that $|f(x)| < 1/|x|$ for all $|x|>1$. How can we show that $f(x)$ is uniformly continuous?

For $f(x)$ to be uniformly continuous, for every $\epsilon>0$, there exists a $\delta>0$ such that $|f(x)-f(y)|< \epsilon$ for every $x,y$ such that $|x-y|< \delta$.

The triangle inequality and the above information imply $|f(x)-f(y)| \leq |f(x)| + |f(y)| \leq 1/|x| + 1/|y|$.

Here, I am stuck, as I cannot relate $1/|x|$ or $1/|y|$ to $|x-y|<\delta$.

Could you please give me a hint on this problem?

Anne Bauval
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RonW
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  • Thank you Anne. While this is not what I looked for, it might answer my question. Let me see if I understand the argument. In bold terms, my function can only attain values in the interval (-1, +1). In addition, it seems that my function converges to a limit L=0. By your theorem then, f(x) must be uniformly continuous. While this is fine, I think it must be uniformly continuous, also because the values of f(x) are confined in the interval (-1, +1), at a first thought. – RonW Aug 31 '23 at 17:24
  • No, boundedness is not sufficient. Think of $\sin(x^2).$ – Anne Bauval Aug 31 '23 at 17:27
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    Very true. Beautiful. Thanks! – RonW Aug 31 '23 at 17:31

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