# Uniformly continuous or not?

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 $$[1,\infty]$$?

• The function $f$ is not defined at $x=0$... – Gibbs Apr 28 at 19:08
• Do you have some tips for how I can find out if it is uniformly continuous? – Mathomat55 Apr 28 at 19:10
• Presumable, you can define $f(0)=0$ to make it continuous on $[0,+\infty).$ – Thomas Andrews Apr 28 at 19:13
• 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. – Thomas Andrews Apr 28 at 19:23
• @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. – 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 $$(0,+\infty)$$