Proof that the sequence $a_n=\frac{3n+2}{n^2+1}$ converges using the Epsilon N proof 
Give a forma Epsilon-N proof that the sequence converges to 0
  $$a_n=\frac{3n+2}{n^2+1}$$
  Hints only!

I find myself still unfamiliarised and uncomfortable with having to deal with the Epsilon-N proof. It might be good to understand the basic it once and for all.
Attempt: 
$$\left|\frac{3n+2}{n^{2}+1}-0\right|< \epsilon$$
At this point I know that for the limit of the sequence to hold, I need 
$$n\leq N$$ but have no idea how to proceed further. 
What question(s) should I be asking myself at this point?
 A: See it as an inequality where you have to solve for $n$.
A: So notice that the sequence is positive for all $n \in \mathbb{N}$, so you can get rid of the absolute value bars.  From there, we have $\displaystyle \frac{3n+2}{n^2+1} < \varepsilon \iff 3n+2 < \varepsilon(n^2 + 1)$.  
So from this we can get an inequality involving a quadratic (this is where your algebra/pre-calculus knowledge will come in handy).  You can show that that inequality is satisfied for all values of $n$ sufficiently large (specifically how large will be in terms of epsilon).  
A: You could also use
$$0\le\frac{3n+2}{n^2+1} \le \frac{3n+3}{n^2-1} = \frac{3(n+1)}{(n+1)(n-1)} = \frac3{n-1}.$$
So now you only need to find an $N$ such that
$$\frac3{n-1}<\epsilon$$
for $n\ge N$.
(We should be a bit careful: This works only for $n\ge1$, since for $n=1$ we have zero in the denominator. But this should not cause much trouble - limit of a sequence is not influenced by changing finite number of terms. We can simply add the condition that $N>1$ and we are fine.) 
