So I supposed to find out if $$f(x)=\frac{1}{1+\ln^2 x}$$ is uniformly continuous on $I=(0,\infty)$ So I have been thinking a lot. Could I say that $f$ is continuous on $[0,1]$ and therefore uniformly continuous here? Or is this not valid, because $\ln$ is not defined at $x=0$? And then say that the derivate is bounded at $[1,\infty]$?

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    $\begingroup$ The function $f$ is not defined at $x=0$... $\endgroup$ – Gibbs Apr 28 at 19:08
  • $\begingroup$ Do you have some tips for how I can find out if it is uniformly continuous? $\endgroup$ – Mathomat55 Apr 28 at 19:10
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    $\begingroup$ Presumable, you can define $f(0)=0$ to make it continuous on $[0,+\infty).$ $\endgroup$ – Thomas Andrews Apr 28 at 19:13
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    $\begingroup$ Yes, you can conclude that $f$ is uniformly continuous on $[0,1]$ after defining $f(0)$ to make $f$ continuous at $0.$ Not sure about the rest of your argument after that, though. $\endgroup$ – Thomas Andrews Apr 28 at 19:23
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    $\begingroup$ @Mathomat55 I think you have problems close to $x=0$. If you extend $f$ so that $f(0)=0$ then you can argue that $f$ is continuous on $[0,1]$, thus uniformly continuous, whereas for $x>1$ the function is differentiable and the derivative is bounded, so it is uniformly continuous on $(1,\infty)$ as well. $\endgroup$ – Gibbs Apr 28 at 19:30

You can extend the function $f$ to a continuous function on $[0,+\infty)$.

Since $\lim_{x\to 0}f(x)$ exists and equal to zero. So we extend and define $f(0)=0$.

We use this theorem:- If $f:[0,+\infty) \to \mathbb{R}$ be continuous on $[0,\infty)$ and $lim_{x\to +\infty}f(x)=0$ then $f$ is uniformly continuous on $[0,\infty)$.

Then observe that, $$\lim_{x\to +\infty}f(x)=0$$. This implies $f$ is uniformly continuous on $[0,+\infty)$ and hence on $(0,+\infty)$

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  • $\begingroup$ Ohh, I have never heard about that theorem before. Thank you! $\endgroup$ – Mathomat55 Apr 28 at 19:37
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    $\begingroup$ You can find a more general version here with proof. $\endgroup$ – Noob mathematician Apr 28 at 19:42

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