Epsilon delta proof question?!?!

Let xn = (3n^2 + 2n − 1)/(n^2 + 2). Find L such that xn → L as n → ∞. Given
Epsilon > 0, find a number N such that |xn − L| < for all n > N.

Let xn = (2n + 1)/√2 + n^4
. Find L such that xn → L as n → ∞. Given
Epsilon > 0, find a number N such that |xn − L| < Epsilon for all n > N


Once I do xn-L i have too many n terms??

The first question is:

Prove that

$\displaystyle \frac{3n^2+2n-1}{n^2+2}=\frac{3n^2+6n-4n-1}{n^2+2}=3-\frac{4n+1}{n^2+2}\rightarrow 3$

So $L=3$

(that is the limit of the sequence is L when for any positive $\epsilon>0$ there exist $N$ ( $N=N(\epsilon)$ ) such that $|x_n-L|<\epsilon$ when
$n>N$

To answer the question you have to solve an inequality

$\displaystyle \left |\frac{3n^2+2n-1}{n^2+2}-3\right |<\epsilon$

that is

$\displaystyle \left |\frac{-4\cdot n-1}{n^2+2}\right |<\epsilon$

$\displaystyle \frac{4n+1}{n^2+2}<\epsilon$

$\displaystyle \epsilon \cdot n^2 +(2\epsilon -4)n-1>0$

$\displaystyle n_{1,2}=\frac{4-2\epsilon \pm \sqrt{16+4\epsilon^2-16\epsilon +4\epsilon}}{2\epsilon}<$

$\displaystyle <\frac{4-2\epsilon + \sqrt{16+4\epsilon^2+16\epsilon }}{2\epsilon}=$

$\displaystyle =\frac{4-2\epsilon+4+2\epsilon}{2\epsilon}=\frac{4}{\epsilon}$
when \epsilon <2

So tha $\displaystyle N=\frac{4}{\epsilon}$ exist for all $\epsilon>0$
so L=3

Why not?