Show that $\frac{n}{n^2-3}$ converges Hi I need help with this epsilon delta proof. The subtraction in the denominator as well as being left with $n$ in multiple places is causing problems.

@Ian has provided a very efficient way to tackle a problem such as this.  I thought it might be instructive to see how tight a bound we can achieve.  To that end, let's write for $n\ge2$
$$\begin{align}
\left|\frac{n}{n^2-3}\right|<\epsilon \implies n&>\frac{1}{2\epsilon}+\frac{1}{2\epsilon}(1+12\epsilon^2)^{1/2} \tag 1\\\\
&=\frac{1}{2\epsilon}+\frac{1}{2\epsilon}\left(1+6\epsilon^2+O\left(\epsilon^4\right)\right)\\\\
&=\frac{1}{\epsilon}+3\epsilon+O\left(\epsilon^3\right)
\end{align}$$
where $(1)$ provides the tightest bound for $N$ as $N=\left\lceil \left(\frac{1}{2\epsilon}+\frac{1}{2\epsilon}(1+12\epsilon^2)^{1/2} \right)\right \rceil$.  

Now, we could also proceed as follows.  We note that $n^2-3>\frac{n^2}{1+\alpha}$, for any $\alpha>0$, provided that $n\ge \left \lceil \sqrt{3\frac{1+\alpha}{\alpha}}\right \rceil$.  For $\alpha =1$, $n^2-3>\frac12 n^2$ for $n\ge \left \lceil \sqrt{6} \right \rceil =3$, which is the inequality imposed by @Ian.  
So, given $\epsilon>0$, we could actually take $N=\max\left(\frac{1+\alpha}{\epsilon},\left \lceil \sqrt{3\frac{1+\alpha}{\alpha}}\right \rceil\right)$, for any $\alpha >0$ and 
$$\left|\frac{n}{n^2-3}\right|<\epsilon$$
whenever $n>N$.
