Prove that $\lim \limits_{n\to\infty}\frac{n}{n^2+1} = 0$ from the definition

You had the right start. We want to show that for any $\epsilon>0$, there exists an integer $N$ such that $\left|\frac{n}{n^2+1}-0\right| <\epsilon$ whenever $n>N$. Or alternately, we want to show that there is an $M$, not necessarily an integer, such that $\left|\frac{n}{n^2+1}-0\right| <\epsilon$ whenever $n>M$.

We solve the problem by producing a suitable $N$ (alternately, $M$).

So let $\epsilon>0$ be given. Note that if $n>0$ then
$$\left|\frac{n}{n^2+1}-0\right|=\frac{n}{n^2+1}<\frac{n}{n^2}=\frac{1}{n}.$$

So if we choose $n$ positive, it is enough to make sure that $\frac{1}{n}<\epsilon$. This inequality is easy to solve for $n$.

The inequality can be rewritten as $n >\frac{1}{\epsilon}$. It will hold if $n > \lceil \frac{1}{\epsilon}\rceil$. So we can choose, for example, $N=\lceil\frac{1}{\epsilon}\rceil$. Here by $\lceil x \rceil$ we mean the smallest integer which is $\ge x$.

Alternately and more simply, if $M=\frac{1}{\epsilon}$, then the desired inequality holds whenever $n>M$.

We have shown that given $\epsilon>0$, if we choose $N=\lceil\frac{1}{\epsilon}\rceil$, then $\left|\frac{n}{n^2+1}-0\right|<\epsilon$ whenever $n>N$. By definition we therefore have $$\lim_{n\to \infty}\frac{n}{n^2+1} =0.$$

Comment: There is no need to solve the inequality $\frac{n}{n^2+1}<\epsilon\:$ exactly. We are not being asked to find the cheapest $N$ such that if $n >N$, then the inequality holds. All we are asked to do is to show that there is such an $N$. It can be very helpful, in this case and elsewhere, to replace our expression $\frac{n}{n^2+1}$ by something which is guaranteed to be larger (but still approaches $0$), and is substantially simpler. In the solution we replaced $\frac{n}{n^2+1}$ by $\frac{1}{n}$. This saved us from the slightly painful task of solving a quadratic inequality.

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?

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 $1$.