Prove that $\lim \limits_{n\to\infty}\frac{n}{n^2+1}  = 0$ from the definition This is a homework question:

Prove, using the definition of a limit, that
  $$\lim_{n\to\infty}\frac{n}{n^2+1}  = 0.$$

Now this is what I have so far but I'm not sure if it is correct:
Let $\epsilon$ be any number, so we need to find an $M$ such that:
$$\left|\frac{n}{n^2 + 1}\right| < \epsilon \text{ whenever }x \gt M.$$
$$ n \lt \epsilon(n^2 + 1) $$
$$n \lt  \epsilon n^2 + \epsilon$$
Now what? 
I am completely clueless on how to do this!

$n<\epsilon n^2 +\epsilon$ if and only if $n^2 - \frac{n}{\epsilon} + 1 > 0$, if and only if $(n - \frac{1}{2\epsilon})^2 + (1 - \frac{1}{4 \epsilon^2}) > 0$ (complete the square)
You should be able to do it from here!  Good luck.
