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I'm trying to prove this from the definition, but i don't now how to proceed.

Davide Giraudo
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Luis Valerin
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    http://math.stackexchange.com/questions/293771/if-p-is-a-non-zero-real-polynomial-then-the-map-x-mapsto-frac1px-is-u – njguliyev Sep 01 '13 at 18:41
  • A continuous function on a compact domain is uniformly continuous. Unfortunately, the entire $(-\infty, \infty)$ is not compact, so you have to do a trick like Prahlad Vaidyanathan suggests in his answer. – Caleb Stanford Sep 01 '13 at 18:54
  • A tip for the future, Luis, is that you should provide more details regarding what you know already about the topic (in this case, about continuity of functions). Since we don't have any background, we might assume (for instance) that you know what a compact set is, or what a Lipschitz function is. So if the answers seem unhelpful or you don't understand them, this is the problem. Cheers – Caleb Stanford Sep 01 '13 at 18:56

2 Answers2

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Your function is differentiable and has a bounded derivative in the whole of $\mathbb R$. You can use the mean value theorem to prove it is in fact Lipschitz.

Mariano Suárez-Álvarez
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Hints :

  1. $\lim_{x\to \pm \infty} f(x) = 0$. Hence, you can find an interval $[-M,M]$ so that it is smaller than $\epsilon/2$ outside that interval.
  2. Any continuous function on $[-M,M]$ is uniformly continuous.
Prahlad Vaidyanathan
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