Is $f(x)=\frac { x }{ { e }^{ x }+1 } $ considered uniformly continuous?

Question

According to a property we proved in my course:

$f(x)$ is uniformly continuous on $(a,b) \Leftrightarrow \exists { L }_{ 1 },{ L }_{ 2 }\in \mathbb R \ s.t.\ lim _{ x\rightarrow a }{ f(x)={ L }_{ 1 },\lim _{ x\rightarrow b }{ f(x)={ L }_{ 2 } } } $

(and to be clearer: $\infty ,-\infty \notin \mathbb R$)

I saw in some book that $\cfrac { x }{ { e }^{ x }+1 }$ is uniformly continuous and it makes sense because it's $\approx f(x) = x$ in the negatives and converges to $0$ on the positives.

Just to make sure - that definition is only when $a,b\in\mathbb R$?

I mean, even $f(x)=x$ is a counter example ...

Answer 1

Yes, this property is only for $a,b\in\mathbb R$ and $f(x)=x$ is a "counterexample" (of cource the correct statement is "...$\iff \color{red}{f \text{ is continuous in } (a,b) \text{ and }} \exists L_1,L_2\in\mathbb R\ldots$").
The function $f(x)=x$ is uniformly continuous in $\mathbb R$ but $\lim_{\pm\infty}f(x)=\pm\infty$.
It is true however, that if $f$ is continuous in $\mathbb R$ and $\lim_{\pm\infty}f(x)\in\mathbb R$ then $f$ is uniformly continuous in $\mathbb R$.
To prove that $f(x)=\dfrac{x}{e^x+1}$ is uniformly continuous in $\mathbb R$ show for example that $f'(x)$ is bounded in $\mathbb R$. Also, $f'(x)$ is uniformly continuous in $\mathbb R$ because $\lim_{x\to{-\infty}}f'(x)=1, \ \lim_{x\to{+\infty}}f'(x)=0.$

Answer 2

Yeah the funciton is uniform continuous, use the implication that Lipschitz continuous implies uniform continuous. Your definition (I really hope that is not your definition) works only when $a,b$ are finite.