If $x_n:=\sqrt{n}$, show that $(x_n)$ satisfies $\lim|x_{n+1}-x_n|=0$, but that it is not a Cauchy sequence.
We see that $$|x_{n+1}-x_n|=|\sqrt{n+1}-\sqrt{n}|=|(\sqrt{n+1}-\sqrt{n})\cdot\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}|=\bigl|\frac{1}{\sqrt{n+1}+\sqrt{n}}\bigr|$$
This seems to simplify nicely, but I am not quite sure where to proceed.
Relevant definitions:
$\lim(x_n) = x$ if and only if for $\varepsilon>0$, $\exists k\in\mathbb{N}$ such that for all $n\geq k$, $|x_n-x|<\varepsilon$.
A sequence $(x_n)$ is a Cauchy sequence if for every $\varepsilon>0$, $\exists h\in\mathbb{N}$ such that for all $n,m\geq h$, $|x_n-x_m|<\varepsilon$.
Take $\epsilon=\frac12$. For any $h\in \mathbb{N}$, choose $n=h$ and $m=4h$. Then $|x_m-x_n|=|\sqrt{4h}-\sqrt{h}|=|2\sqrt{h}-\sqrt{h}|=\sqrt{h} \ge 1$ which contradicts the definition of Cauchy sequence for above $\epsilon$.