Lipschitz function and incomplete range Find a Lipschitz function $f:\mathbb{R} \rightarrow \mathbb{R}$ that has incomplete range.

What is incomplete range of a function? Does it has to do with completeness of a metric space?

Any help would be appreciated.
If by incomplete range we mean that the image $f(\Bbb R)$ of $f:\Bbb R \to \Bbb R$ is not a complete subset of $\Bbb R$, then the function

$f(x) = \tanh x \tag 1$

provides an example.  It is everywhere continuously differentiable, which is sufficient for it to be
Lipschitz; but

$f(\Bbb R) = \tanh(\Bbb R) = (-1, 1), \tag 2$

which is incomplete as there are sequences $\alpha_i \in (-1, 1)$ such that

$\alpha_i \to -1, \; \text{or} \; \alpha_i \to 1 \; \text{as} \; i \to \infty, \tag 3$

but

$-1, \; 1 \notin (-1,  1). \tag 4$
If by incomplete range we mean that the image $f(\Bbb R)$ of $f:\Bbb R \to \Bbb R$ is not a complete subset of $\Bbb R$, then the function

$f(x) = \tanh x \tag 1$

provides an example.  It is everywhere continuously differentiable, which is sufficient for it to be
Lipschitz; but

$f(\Bbb R) = \tanh(\Bbb R) = (-1, 1), \tag 2$

which is incomplete as there are sequences $\alpha_i \in (-1, 1)$ such that

$\alpha_i \to -1, \; \text{or} \; \alpha_i \to 1 \; \text{as} \; i \to \infty, \tag 3$

but

$-1, \; 1 \notin (-1,  1). \tag 4$