Given a sequence $a_n =\frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$, prove that $\lim_{n \to \infty} a_{n} = g$. The given Sequence is $a_n = \frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5}$.
I showed that the Sequence $a_n$ converges towards a Value $g = \frac{1}{3}$.
How do I determine for each $\epsilon > 0$ an $n_0$ so that: $|a_n - g| < \epsilon \ \forall n > n_0$?
This is what I tried so far:
$\begin{array}{rcl}
\left|a_n - g\right| & < & \epsilon \\
\left|\frac{n^2 - 3n + (-1)^n}{3n^2 - 7n + 5} - \frac{1}{3}\right| & < & \epsilon \\
\left|\frac{3n^2 - 9n + 3 \cdot (-1)^n}{9n^2 - 21n + 15} - \frac{3n^2 - 7n + 5}{9n^2 - 21n + 15}\right| & < & \epsilon \\
\left|\frac{-2n - 5 + 3 \cdot (-1)^n}{9n^2 - 21n + 15}\right| & < & \epsilon \ \text{ Denominator > 0, Numerator < 0 } \forall n \in \mathbf{N} \\
\frac{2n + 5 - 3 \cdot (-1)^n}{9n^2 - 21n + 15} \leq \frac{2n + 5 + 3}{9n^2 - 21n + 15} = \frac{2n + 8}{9n^2 - 21n + 15} & < & \epsilon \\
\end{array}$
Are my Steps correct so far?
How do I proceed from here?
 A: What you have done is correct. You need $\frac {2n+8} {3n(3n-7)+15} <\epsilon$. 
One way to achieve this is to make $3n-7 >\frac 1 {\epsilon}$. $\cdots$ (1).
Now we need $2n+8 <3n+15\epsilon$. For this we need 
$n>8-15\epsilon$. Hence $n >8-15\epsilon$ and $n>\frac 1  3 (7+\frac 1 {\epsilon})$ is good enough. Choose $n_0$ to be greater than the maximum of  $\frac 1  3 (7+\frac 1 {\epsilon})$ and $8-15\epsilon$
A: Your work is correct. You can continue with more inequalities. For $n>8$, we have $2n+8<3n$ and $9n^2-21n+15>n^2$ since $8x^2-21x+15=8\left(x-\frac{21}{16}\right)^2+\frac{39}{32}>0$ for all $x\in\mathbb{R}$. Hence,
$$\frac{2n+8}{9n^2-21n+15}<\frac{3n}{n^2}=\frac{3}{n} $$
Now you just need to find $n_1$ such that when $n>n_1$, we have $\frac{3}{n}<\varepsilon$. This shouldn't be too hard. Since we also want to incorporate $n>8$, we choose $n_0=\max(n_1,8)$.
A: This follows directly from the definition of a Limit of a sequence. if $a_n$ converges to $g$, then $lim_{n\to \infty} a_n =g$, such that $\forall \epsilon >0 \exists n_0 \in \mathbb{N} \text{ such that } \forall n\ge n_0 : |a_n -g|<\epsilon$
