Proving a function is uniformly continous on $(0, \infty).$

Question

I'm trying to prove the following function is uniformly continous on $(0, \infty)$.

$$ f(x) = \frac{\sin(x)}{e^x - e^{-x}} $$ I didn't make much progress at all using the defintion of uniform continuity, and so tried to use the bounded derivative test. That amounts to proving the following function is bounded.

$$ f'(x) = \frac{\cos(x)(e^x - e^{-x}) - \sin(x)(e^x + e^{-x})}{(e^x - e^{-x})^2} $$

However I've been stuck on this also. I've also not been able to do it using other methods like $f(x)$ preserving cauchyness.

Any help at all on this question would be very much appreciated.

Answer 1

This is a long comment to elucidate the details of the A by @Jose Carlos Santos.

1.Extend $f$ to the domain $[0,\infty)$ with $f(0)=1/2.$ Then $f$ is continuous on $[0,\infty)$ so $f$ is uniformly continuous on $[0,a]$ for any $a\in \Bbb R^+.$

2. $\lim_{x\to \infty}f(x)=0$ because if $x>0$ then $|\sin x|\le 1$ and $e^x-e^{-x}>e^x-1>0$ so $|f(x)|\le \frac {1}{|e^x-e^{-x}|}=\frac {1}{e^x-e^{-x}}<\frac {1}{e^x-1}.$

3. Given $\epsilon>0$: By 2. take $a>0$ such that $x\ge a\implies |f(x)|<\epsilon /3.$ Then by 1. take $\delta >0$ such that $(x,y\in [0,a]\land |x-y|<\delta)\implies |f(x)-f(y)|<\epsilon /3.$

Now if $0\le x\le y<x+\delta$ then

(i) If $0\le x\le y\le a$ then $|f(x)-f(y)|<\epsilon/3<\epsilon. $

(ii) If $a\le x\le y$ then $|f(x)-f(y)|\le |f(x)|+|f(y)|<\epsilon/3+\epsilon/3<\epsilon.$

(iii) If $0\le x\le a\le y$ then $a,x \in [0,a]$ with $|a-x |<\delta$ [... because $0\le a-x\le y-x<\delta$ ...] so $|f(x)-f(a)|\le \epsilon /3.$ And each of $|f(a)|,|f(y)|$ is $<\epsilon/3.$ Therefore $$|f(x)-f(y)|\le |f(x)-f(a)|+|f(a)-f(y)|\le $$ $$\le|f(x)-f(a)|+|f(a)|+|f(y)|<$$ $$<\epsilon/3 +\epsilon/3+\epsilon/3.$$

Answer 2

Take $a>0$. Since $\lim_{x\to0}f(x)=\frac12$ (the only thing that matters here is that this limit exists in $\Bbb R$) and $f$ is continuous, the restriction of $f$ to $(0,a]$ is uniformly continuous. And it follows from your computations that it is uniformly continuous on $[a,\infty)$. Therefore, $f$ is uniformly continuous.