Proving that uniformly continuous functions respect Cauchy sequences in their domain

Question

I'm having a hard time coming up with a valid Cauchy sequence that fits this criteria. I am unsure of a solid approach to determine such a Cauchy sequence, besides trial and error with something that appears to affect $\sin{\frac{1}{x}}$.

Let $A = (0, \infty)$. The function $f(x) = \sin{\frac{1}{x}}$ is continuous on $A$. Show that for this function $f$ there is a Cauchy sequence $\{x_{n}\}$ in its domain such that $\{f(x_{n})\}$ is not a Cauchy sequence in $\mathbb{R}$.

Answer 1

Let $x_n = \frac{1}{\frac{\pi}{2}+n\pi}$ then $x_n $ is Cauchy sequence but $f(x_n )=(-1)^n $ is not.