Uniformly continuous or not?

Question

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 $[1,\infty]$?

Answer 1

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 $(0,+\infty)$