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!

You could harvest from
$\frac{n}{n^2+1} = \frac{1}{n+\frac{1}{n}}$ for any $n$ in $\mathbb{R}^\star$, and for any $\varepsilon>0,n>0$ then $\frac{1}{n+\frac{1}{n}}<\varepsilon \Leftrightarrow n + \frac{1}{n} > \frac{1}{\varepsilon}$.
On the range $\left[1;+\infty\right[$ then $\frac{1}{n}$ is upper-bounded by $1$.
