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!

First, $\epsilon$ should not be "any number", it should be "any positive number."
Now, you are on the right track. What do you need in order for $\frac{n}{n^2+1}$ to be smaller than $\epsilon$? You need $n\lt \epsilon n^2 + \epsilon$. This is equivalent to requiring
$$\epsilon n^2 - n + \epsilon \gt 0.$$
You want to find out for what values of $n$ this is true. This is a quadratic inequality: you first solve
$$\epsilon n^2 - n + \epsilon = 0,$$
and then you use the solution to determine where the quadratic is positive, and where it is negative. The answer will, of course, depend on $\epsilon$.
Using the quadratic formula, we have that
$$\epsilon n^2 - n + \epsilon = 0$$
has solutions
$$n = \frac{1 + \sqrt{1-4\epsilon^2}}{2\epsilon}, \quad n= \frac{1-\sqrt{1-4\epsilon^2}}{2\epsilon}.$$
That is,
$$\epsilon n^2 - n + \epsilon = \epsilon\left( n - \frac{1-\sqrt{1-4\epsilon^2}}{2\epsilon}\right)\left(n - \frac{1+\sqrt{4\epsilon^2}}{2\epsilon}\right).$$
Now, we can assume that $\epsilon\lt \frac{1}{2}$, so that $4\epsilon^2\lt 1$ (if it works for all small enough $\epsilon$, then it works for all $\epsilon$. Since $\epsilon\gt 0$, then the quadratic is positive if $n$ is smaller than the smallest of the roots, or if $n$ is larger than the larger of the two roots. The larger root is $\displaystyle \frac{1 + \sqrt{1-4\epsilon^2}}{2\epsilon}$. So if
$$n \gt \frac{1+\sqrt{1-4\epsilon^2}}{2\epsilon},$$
then
$$\epsilon n^2 -n + \epsilon \gt 0.$$
Can you finish it up from here?
