How to explicitly find a natural number N that satisfies the statement in the definition of the limit? For example, in this question:
$\lim _{x\to \infty \:}\left(\frac{x^2}{x^2+2x}\right)=1$
For every epsilon > 0, I am asked to find explicitly a natural number N which
satisfies the statement in the definition of the limit.
What does find explicitly $N$ means? Do I just prove it as usual?
$|\frac{x^2}{x^2+2x} -1|< Ɛ$
Continue with this I got:
$n+2 > \frac{2}{Ɛ}$
and then I take:
$ N = 2[\frac{1}{Ɛ}] + 1$
then since $n >= N$:
$n+2 >= 2[\frac{1}{Ɛ}] + 1 + 2$
therefore, $N = 2[\frac{1}{Ɛ}] + 3$, did I just found my $N$ or is it asking for something numerical? I am confused?
 A: Per back and forth with OP, this answer is intended to eliminate any confusion around the problem.  The answer will (mostly) use analysis already done by the OP.


For example, in this question:


$\lim _{x\to \infty \:}\left(\frac{x^2}{x^2+2x}\right)=1$


For every epsilon > 0, I am asked to find explicitly a natural number N which
satisfies the statement in the definition of the limit.

For all positive integers $n$, let $a_n$ be defined as
$\left(\frac{n^2}{n^2+2n}\right).$
There are two ways that the problem might be interpreted.  The narrow way is that given $\epsilon > 0$, you are being asked to find $N \in \mathbb{Z^+}$ such that $|a_N - 1| < \epsilon.$
The alternative intrepretation, which I will adopt, and which might not have been intended is that you are to find $N \in \mathbb{Z^+}$ such that for every element $a_n$ in the subsequence
$\{a_{N}, a_{(N+1)}, a_{(N+2)}, \cdots \}$ you have that 
$|a_n - 1| < \epsilon.$
The reason that I have chosen the 2nd interpretation, which is moderately more onerous, is that this interpretation represents the standard definition of a sequence converging to a limit.
That is, normally, you show that a sequence $\langle a_n\rangle$ converges to a limit $L$ by showing that for any $\epsilon > 0$ there exists a positive integer $N$, such that for all positive integers $n \geq N$ (or $n > N$), $|a_n - L| < \epsilon.$
Hijacking the analysis already done by the OP: 
it is clear that each element $a_n$ will be between $0$ and $1$. 
Therefore, the first objective is to find an $N$ such that
$$1 - a_N < \epsilon \implies \tag{Constraint-1}$$
$1 - \frac{N^2}{N^2 + 2N} < \epsilon \implies$
$\frac{N^2 + 2N - N^2}{N^2 + 2N} < \epsilon \implies$
$\frac{2N}{N^2 + 2N} < \epsilon \implies$
$\frac{2}{N+2} < \epsilon \implies$
$\frac{N+2}{2} > \frac{1}{\epsilon} \implies$
$$(N+2) > \frac{2}{\epsilon}.\tag{Constraint-2}$$
What you actually have is that if Constraint-1 is satisfied for a specific positive integer $N$, then Constraint-2 is also satisfied.  However, what you want is the reverse.
Notice which way the $\implies$ symbols are directed.  However, if you examine the actual analysis step-by-step, you will see that each step has a (hidden) bidirectional implication.  That is a given step is true $\iff$ the next step is true.
This means that if Constraint-2 is satisfied for a specific positive integer $N$, then Constraint-1 will also be satisfied.
It further means that if $n$ is any positive integer that satisfies Constraint-2, then $n$ will also satisfy Constraint-1.
Therefore, for a given fixed $\epsilon > 0$, specify $N$ by Constraint-2.  This implies that $a_N$ satisfies Constraint-1.
But there is more.
Suppose that $n$ is any positive integer such that $n > N$.
Then, $(n+2) > (N+2).$ 
Therefore, $n$ also satisfies Constraint-2. 
Therefore, $a_n$ also satisfies Constraint-1.
Thus, Constraint-2 provides the specification for $N$, based on $\epsilon$ such that each element in 
$\{a_N, a_{(N+1)}, a_{(N+2)}, \cdots \}$ satisfies Constraint-1.
