Equivalent alternative to delta-epsilon formulation of limit? Is $\lim_{n \to \infty} x_n = L$ same as:

$\forall n \in \mathbb N,
     \exists \ \varepsilon_n > 0 $ so that:  $$  \ |x_n - L| \leq
     \varepsilon_n$$ and $$\varepsilon_n \to
     0, \ \text{when } n \to \infty$$

I know the second assumes and uses the definition of limit again, but my question is if they are equivalent.
Thanks.
 A: You can take $\epsilon_n = 2|x_n-L|$, since you assume $\lim_{n \to \infty} x_n = L$ for every $\mu > 0$ there exists $N>0$ such that $$|x_n-L|< \mu / 2 \qquad \forall n \geq N $$ and so we 
  have 
$$|\epsilon_n| = 2|x_n-L|< \mu \qquad \forall n \geq N$$
this shows that $\lim_{n \to \infty} \epsilon_n = 0$.
A: The better way to formulate something like your description is to first rewrite the definition of the limit as
$$(\exists \delta : \mathbb{R}_+ \times \mathbb{R} \to \mathbb{R}_+)(\forall \varepsilon > 0,x \in \mathbb{R}) \, |x-c|<\delta(\varepsilon,x) \Rightarrow |f(x)-L|<\varepsilon.$$
The function whose existence is implied by this statement is called the modulus of convergence. Now suppose $\delta_1$ is a modulus of convergence and $\delta_2 \leq \delta_1$. Then $\delta_2$ is also a modulus of convergence. 
Therefore we can assume that the modulus of convergence is strictly increasing and continuous, by taking an arbitrary modulus of convergence and decreasing it. This means the modulus of convergence has a continuous inverse, which I will call $\varepsilon : \mathbb{R}_+ \times \mathbb{R} \to \mathbb{R}_+$. 
What we have said is just:
$$(\exists \varepsilon : \mathbb{R}_+ \times \mathbb{R} \to \mathbb{R}_+)(\forall \delta > 0,x \in \mathbb{R}) \, |x-c|<\delta \Rightarrow |f(x)-L|<\varepsilon(\delta,x).$$
But the original statement also implies that
$$(\forall x \in \mathbb{R}) \, \lim_{\delta \to 0^+} \varepsilon(\delta,x) = 0$$
because $\varepsilon(\delta,x)$ is the inverse of $\delta(\varepsilon,x)$ which has the same property.
Everything here should go through just fine in the opposite direction: if we assume $\lim_{\delta \to 0^+} \varepsilon(\delta,x) = 0$ then we can invert it (by increasing it if necessary) to get a function $\delta$ defined for arbitrarily small $\varepsilon$, which is what is required for convergence.
A: I can't make sense of the  "as ...  tends to infinity/zero"  parts of your formulation.  The great thing about the definition of limits is, that they are totally finitary. No infinity is involved. 
So my answer is: No the definitions do not make sense. 
