use $N-\varepsilon $ language to prove the statement $\forall \epsilon \in (0,1) \ \exists N \in \mathbb{N^+} :\ \forall n\ge N:\ |x_n-a|\le 2\varepsilon \Leftrightarrow  \forall\varepsilon_1>0,\exists N \in \mathbb{N^+},n\ge N,|x_n-a|<\varepsilon_1$ 
how to manipulate $N-\varepsilon $ language to prove above statement? 
 A: A general understanding of the statements we're dealing with is a good thing:
The first statement is telling: given any real number $\epsilon$ between zero and one, we can find (that's the core of the statement) a natural number $N$ so that the distance $|x_{n\geq N} - a|$ between the sequence's elements $x_{n\geq N}$ starting at $n=N$ and the limit $a$ oft the sequence $x_n$ is smaller than the double of the given real number $\epsilon$.
The second statement is telling: given any real number $\epsilon_1$ greater than zero,  we can find a natural number $N$ so that the distance $|x_{n\geq N} - a|$ between the sequence's elements $x_{n\geq N}$ starting at $n=N$ and the limit $a$ of the sequence $x_n$ is smaller than the given real number $\epsilon_1$.

What needs to be done in general
If you want to prove the equivalence, then: 
(a)  assume that the first statement is true and prove the second statement 
(b)  assume the second statement is true and prove the first statement.

The (a) direction or the $\implies$ direction
(a) assume for any given $0<\epsilon<1$ we can find a natural number $N_0$ so that  $|x_{n\geq N_0} - a|<2\epsilon$, that is assume the first statement is true. To prove the second statement, we need to find a number $N_1$ for any given $\epsilon_1 >0$, so that $|x_{n\geq N_1} - a|<\epsilon_1$. 
First we prove the following implication
$\forall\epsilon \in (0,1),\exists N_0 \in \mathrm{N^+},|x_{n\ge N_0}-a|\le 2\epsilon \implies  \forall\epsilon_1>0,\exists N_1 \in \mathrm{N^+},|x_{n\ge N_1}-a|<\epsilon_1$
What is the premise? What is the assumption? 
Assume we can find a number $N_0$ so that the condition or the inequality $|x_{n\geq N_0} - a|<2\epsilon$ is true for any given $0<\epsilon<1$. 
What is the conclusion or entailment?
The task is to find or show that we can find a number $N_1$ so that the inequality or the condition $|x_{n\geq N_1} - a|<\epsilon_1$ is true for any given $\epsilon_1 > 0$. 
How do we mathematicize or how do we use the assumption to get to the conclusion? 
Now if we're given an $0<\epsilon_1^x<2$, then it follows directly from the assumption, that  $|x_{n\geq N_1^x} - a|<\epsilon_1^x$ and we can find an $N_1^x$ (in this particular case of $0<\epsilon_1^x<2$ we can just choose or set $N_1^x=N_0$) so that the condition  $|x_{n\geq N_1^x} - a|<\epsilon_1^x$ for any given $0<\epsilon_1^x<2$ is satisfied. 
In the case we're given any $\epsilon_1^y \geq 2$ and asked to find an $N_1^y$ so that $|x_{n\geq N_1^y} - a|<\epsilon_1^y$ is true, we note that $|x_{n\geq N_1^y} - a|<\epsilon_1^x<\epsilon_1^y$. This shows that we can find such an $N_1^y$ that makes $|x_{n\geq N_1^y} - a|<\epsilon_1^y$ a true condition.
With being able to find a number $N_1^x$ so that $|x_{n\geq N_1^x} - a|<\epsilon_1^x$ is true and a number $N_1^y$ so that $|x_{n\geq N_1^y} - a|<\epsilon_1^y$ is also true for any given $0<\epsilon_1^x<2$ and for any given $\epsilon_1^y \geq 2$, we have proved now that we can find an $N_1$ so that $|x_{n\geq N_1} - a|<\epsilon_1$ is true for any given $\epsilon_1>0$.

The (b) direction or the $\impliedby$ direction
(b) assume the second statement is true and imply the first statement. Prove the following implication:
$\forall\epsilon \in (0,1),\exists N_0 \in \mathrm{N^+},|x_{n\ge N_0}-a|\le 2\epsilon \impliedby  \forall\epsilon_1>0,\exists N_1 \in \mathrm{N^+},|x_{n\ge N_1}-a|<\epsilon_1$
If $\forall\epsilon_1>0,\exists N_1 \in \mathrm{N^+},|x_{n\ge N_1}-a|<\epsilon_1$, then clearly $\forall\epsilon \in (0,1),\exists N_0 \in \mathrm{N^+},|x_{n\ge N_0}-a|\le 2\epsilon$, because $2\epsilon > 0$.

After $\implies$ and $\impliedby$, then you have shown the equivalence $\iff$, because showing that $(A \implies B)$ and showing $ (A \impliedby B)$ for two statements $A$ and $B$ is equivalent to saying that $A \iff B$.
A: Given the first statement, we wish to prove the second. We start with $\epsilon_1>0$ and wish to find $N$ so that $|a_n-a|<\epsilon_1$ for all $n\ge N$. Set $\epsilon = \epsilon_1/2$. The first statement provides us an $N$ so that
$$|a_n-a|<2\epsilon = 2\cdot\frac{\epsilon_1}2 = \epsilon_1$$
for all $n\ge N$. So that $N$ will work. [You can call the $N$ from the first statement $N_0$, if that makes it clearer to you.]
Can you do the other direction?
A: Can I do like this?:
$$\begin{aligned}
left\Rightarrow right:& \forall\varepsilon \in (0,1),\exists N \in \mathrm{N^+},n\ge N,|x_n-a|\le 2\varepsilon
\\& \frac{\varepsilon_{1}}{2}=\varepsilon,\Rightarrow \forall \varepsilon= \frac{\varepsilon_{1}}{2} \in (0,\frac{1}{2} ] \Leftrightarrow \forall \varepsilon_1 \in (0,1] ,\exists N_2 \in \mathrm{N^+},n\ge N_2,|x_n-a|\le \varepsilon_1
\\& \varepsilon=\frac{2}{\pi}\arctan(\varepsilon_2) \in (\frac{1}{2},1) \\&\Leftrightarrow \forall\varepsilon_2 \in(1,+\infty),\exists N_3 \in \mathrm{N^+},n\ge N_3,|x_n-a|\le 2\varepsilon= \frac{4}{\pi}\arctan(\varepsilon_2)\le \varepsilon_2
\\&\varepsilon_3=\left\{\begin{aligned}
& \varepsilon_1=2\varepsilon & \varepsilon \in (0,\frac{1}{2 } ],
\\& \varepsilon_2=\tan(\frac{\pi}{2}\varepsilon) & \varepsilon \in (\frac{1}{2} ,1).
\end{aligned}\right., N_4=\max\{N_3,N_2\}
\\&\Rightarrow \forall \varepsilon_3 >0 , \exists N_4 \in \mathrm{N^+},n\ge N_4 ,|x_n-a|\le \varepsilon_3 \Leftrightarrow \lim_{n\to\infty}{x_n}=a
\\right \Rightarrow left:&\forall \varepsilon >0 , \exists N \in \mathrm{N^+},n\ge N ,|x_n-a|\le \varepsilon, 2\varepsilon_1=\varepsilon >0, \varepsilon_1=\frac{\varepsilon}{2}>0 
\\& \forall \varepsilon_1>0 ,\exists N_1 \in \mathrm{N^+},n\ge N_1, |x_n-a|\le 2\varepsilon_1.
\\&  \varepsilon_2 \in (0,1)  \subset (0,+ \infty) \ni \varepsilon_1 \Rightarrow \forall \varepsilon_2 \in(0,1) ,\exists N_2 \in \mathrm{N^+},n\ge N_2 ,|x_n-a|\le 2\varepsilon_2.
\\&\square
\end{aligned}$$
