Does a bounded continuous function map Cauchy sequences to Cauchy sequences?

Question

I only ever see the example of $f:(0,1]\rightarrow \mathbb{R}$ where $f(x)=\frac{1}{x}$as that of a continuous function that does not map Cauchy sequences to Cauchy sequences.

Are there examples of bounded continuous functions that do not map Cauchy sequences to Cauchy sequences? I am not looking for wild topologies here, but more so if this is possible for, say, a holomorphic function on the unit disk.

Thanks!

Answer 1

$f(x)=\sin\frac1x$ on $(0,1]$. For the disk you just need a bounded holomorphic function on the disk that's discontinuous on the boundary. Such functions do exist, as answers to Continuous Extension of a Bounded Holomorphic Function on the unit disk? show, but it's harder to produce an explicit example.

Answer 2

$\sin(1/x)$ maps the sequence $x_n=\frac1n$ to $\sin(n)$ which is definitely not Cauchy.