Cauchy criterion for a function limit (${\lim }_{h\to 0^{+}}F(h)\text{ exists}⟺\forall \epsilon >0\exists \delta >0:d(F(h),F(k))<\epsilon (0<h,k<\delta )$)

For a function $F:(0,1)\to Z$ into a complete metric space, the displayed pairwise condition is equivalent to existence of a limit as $h\to 0^{+}$. Necessity follows from the triangle inequality. For sufficiency, the values $F(1/(n+1))$ form a Cauchy sequence, hence converge to some $z\in Z$. Given $\epsilon >0$, use the pairwise condition with $\epsilon /2$ and then send $n\to \infty$ in $d(F(h),F(1/(n+1)))<\epsilon /2$. This gives $d(F(h),z)\le \epsilon /2<\epsilon$ for every sufficiently small positive $h$.