Let $\xi$ be a limit point of the domain $D$ of $f$. Then $f$ _satisfies the Cauchy condition_ for $x\to\xi$ if for any given $\epsilon>0$ there is a neighborhood $U$ of $\xi$ such that $$|f(x)-f(y)|<\epsilon\qquad\forall x, \> y\in\dot U\ .$$
If $D\subset{\mathbb R}$ is a finite interval, and $\xi=\sup D$ then the Cauchy condition means that for any $\epsilon>0$ we can find a $\delta>0$ such that $$|f(x)-f(y)|<\epsilon\qquad\forall\> x, \>y\in\ ]\xi-\delta, \ \xi[\ .$$ If $D=[a,\infty[\ $ for some $a\in{\mathbb R}$ then the Cauchy condition makes sense for $x\to\infty$, and reads as follows: For any given $\epsilon>0$ there is an $M\geq a$ such that $$|f(x)-f(y)|<\epsilon\qquad\forall \>x, \>y>M\ .$$ One has the following
**Theorem:** If $f$ satisfies the Cauchy condition for $x\to\xi$ then the limit $\lim_{x\to\xi}f(x)$ exists.