I have this problem: is the function
$$f(x)=\frac{\ln(x+1)}x$$
uniformly continuous in $\;[0,\infty)\;$ ?
First, I think I could say that zero is not even in domain of function, but I can show it is a movable right discontinuity since $\;f(x)\to 1\;$ when $\;x\to 0^+\;$ , so I suppose we can have the function defined in there (is htis true?)
Now, for the uniform continuity: is possible to show that
$$f'(x)=\frac{ x-(x+1)\ln(x+1)}{x^2(x+1)}<0\;\;in\;\; [0,1]\;$$
and the function decreases in monotonically, and we also have that using LHospital
$$f'(x)\xrightarrow[x\to 0^+]{}-\frac12$$
$$f'(x)\xrightarrow[x\to \infty]{}0$$
So I think first derivative is bounded and then I get uniformly continuity.
Is the above right? Anything to say about this I'll thank very much
You have $f'(x) = {x - (x+1)\log(x+1) \over x^2(x+1)}$. It is clear that $f'(x) \to 0$ as $ x \to \infty$. Using L'Hôpital we have $f'(x) \to - {1 \over 2}$ as $x \downarrow 0$. Since $f'$ is continuous on $(0,\infty)$, we see that $f'$ is bounded, that is $|f'(x)| \le M$ for some $M$.
Then the mean value theorem gives $|f(x)-f(y)| = |f'(\xi)| |x-y| \le M |x-y|$, and so $f$ is Lipschitz continuous.
In particular, given $\epsilon>0$, if $|x-y|< {1 \over M}\epsilon$, then $|f(x)-f(y)| < \epsilon$.