# How to prove that the following statements are equivalent: $\lim_{n\to \infty} a_n = a$ and $\lim_{x\to 0} \bar{\alpha}(x) = a$

In an assignment for the course Real-Analysis I need to proof that that the following statements are equivalent:

1. $$\lim_{n\to \infty} a_n = a$$
2. $$\lim_{x\to 0} \bar{\alpha}(x) = a$$

Given that $$(W, d_w)$$ is a metric space and $$\alpha: \mathbb{N} \longrightarrow W$$ and $$\alpha(n) = a_n$$ a row in $$W$$. We can use the function $$\bar{\alpha}:\mathbb{R} \supset \longrightarrow W$$ with:

$$Dom(\bar{\alpha}) = \{\frac{1}{n+1} : n \in \mathbb{N}\} \subset \mathbb{R}$$

and

$$\bar{\alpha}(x) = \alpha (\frac{1}{x} -1), x \in Dom(\bar{\alpha})$$

Let $$a \in W$$ and prove that the following statements are equivalent:

1. $$\lim_{n\to \infty} a_n = a$$
2. $$\lim_{x\to 0} \bar{\alpha}(x) = a$$

However, me and my fellow students have been trying to figure it out for some time now, but we can't...

Let's write $$x=\frac{1}{n+1}~~~n \in \mathbb{N}$$. $$\bar{\alpha}(\frac{1}{n+1})=\alpha(n)=a_n~~~\forall n$$.
$$1. \Rightarrow 2.$$ For given $$\epsilon >0~~~\exists N \in \mathbb{N}$$ such that $$d_w(a_n,a)<\epsilon$$ for all $$n >N$$. Then for $$\frac{1}{n+1}< \frac{1}{N+1}$$ ($$|x|<\delta$$),
$$d_w(\bar\alpha(\frac{1}{n+1}),a)\leq d_w(\bar\alpha(\frac{1}{n+1}),a_n)+d_w(a_n,a)<\epsilon$$
$$2. \Rightarrow 1.$$ Given $$\epsilon >0~~~\exists\delta>0$$ such that $$d_w(\bar\alpha(\frac{1}{n+1}),a)<\epsilon$$ whenever $$\frac{1}{n+1}<\frac{1}{M+1}<\delta$$ (for some large $$M \in \mathbb{N}$$). Then for $$n >M$$, $$d_w(a_n,a)\leq d_w(\bar\alpha(\frac{1}{n+1}),a_n)+d_w(\bar\alpha(\frac{1}{n+1}),a)<\epsilon$$