Basic Cauchy sequence convergence question Let $f: ]0,1[ \to \mathbb{R}$ be a function. Suppose that for every sequence $(\epsilon_n)_n$ in $]0,1[$ with $\epsilon_n \searrow 0$ we have that $(f(\epsilon_n))_n$ is a Cauchy sequence. Can we deduce that 
$$\lim_{\epsilon \to 0} f(\epsilon)$$
exists? 
We kno that $(f(\epsilon_n))$ converges  for all $n$ but not necessarily to the same limit otherwise we would be done.
 A: If you had two sequences, $(\epsilon_n)$ and $(\epsilon'_n)$ producing different limits for $(f(\epsilon_n))$ and $(f(\epsilon'_n))$ then, by suitably interleaving the terms form both sequences $(\epsilon_n)$ and $(\epsilon'_n)$, you would get a decreasing sequence $(\delta_n)$, converging to $0$, for which $(f(\delta_n))$ isn't a Cauchy sequence, because its terms get close to both of the limits for $(f(\epsilon_n))$ and $(f(\epsilon'_n))$.
A: Take two sequences $\varepsilon^i_n$ going to zero and define $\varepsilon^3_{2n} = \varepsilon^1_n$ and $\varepsilon^3_{2n+1} = \varepsilon^2_n$. Then $\varepsilon^3_n$ is going to zero too. $f(\varepsilon^3_n)$ is then converging, and by extracting, its limit is equals to those of $f(\varepsilon^1_n)$ and of $f(\varepsilon^2_n)$. Thus the two previous limits are equal.
Edit : to construct a decreasing sequence out of $\varepsilon^1$ and $\varepsilon_2$ which are decreasing sequences, we construct it inductively :
$\varepsilon^3_0 = \varepsilon^1_0$,
$\varepsilon^3_{2n+1} = \sup\{\varepsilon^2_k \text{ s.t } \varepsilon^2_k < \varepsilon^3_{2n} \}$, $\varepsilon^3_{2n} = \sup\{\varepsilon^1_k \text{ s.t } \varepsilon^1_k < \varepsilon^3_{2n-1} \}$. I think that works.
