Proving $\lim_{x \to \infty}\frac{2|x|}{x+1} = 2$ using definition I'm trying to prove this limit here

Prove $\lim_{x \to \infty}\frac{2|x|}{x+1} = 2$, using epsilon-delta or sequence limit definition

Here's the answer I've come up so far
Let $\epsilon > 0 $, by Archimedian Property, then exist $m \in  N$ such that
Consider $\epsilon ' = \frac{1}{2}\epsilon$, it's clear that $\epsilon ' > 0$ then we get $\frac{1}{m} < \epsilon '$
Then, for every $x ≥ m$, we get
$|\frac{2|x|}{x+1} - 2| = |\frac{2|x|-2x-2}{x+1}| = |\frac{2x-2x-2}{x+1} | = |\frac{-2}{x+1} | = \frac{2}{x+1} ≤ \frac {2}{x} \leq \frac{2}{m} < 2\epsilon ' = \epsilon $
limit proven.
Is this correct? also it seems there is another way to prove this? like using epsilon-delta definition.
Any insight would really help, thanks beforehand.
 A: Although the  OP's intent can be understood, it is advised that they carefully "'dot their i's" and "cross their t's" to logically nail it down and they can then compare their technique to the one given here.
The OP should understand that

When taking a limit out to infinity, such as
$\quad \displaystyle \lim_{x \to +\infty} f(x) = L$
that the  $\delta \gt 0$ 'bit' has a different interpretation (see the last section).

Set $f(x) = \frac{2|x|}{x+1}$. It is easily shown (use inequality algebra) that
$\quad \displaystyle f\bigr(\,[0,+\infty)\,\bigr) \subset [0, 2]$
So we now only have to address a challenge $\varepsilon$ satisfying  $0 \lt \varepsilon \lt 2$.
For $x \gt 0$
$\quad f(x) \ge 2-\varepsilon \text{ iff }$
$\quad \quad  2x \ge 2x + 2 -\varepsilon x  - \varepsilon \text{ iff }$
$\quad \quad  \varepsilon x  \ge 2 - \varepsilon \text{ iff }$
$\quad \quad  x  \ge \frac{2 - \varepsilon}{\varepsilon}$
Setting $d = \frac{2 - \varepsilon}{\varepsilon}$ we can now write as true
$\quad \displaystyle f\bigr(\,[d,+\infty)\,\bigr) \subset [2 - \varepsilon, 2 + \varepsilon]$
and so
$\quad \displaystyle \lim_{x \to \infty}\frac{2|x|}{x+1} = 2$

The reader can review the definition
$\quad$ Limits at infinity
This definition uses strict inequalities and the limit control variable is designated with the letter $c$, but the above is an equivalent formulation.
Heck, you could even use $\delta$ rather that $c$ or $d$, but that would bring a frown to the face of some mathematicians.
