$\varepsilon$-$N$ proof question 
Use the definition of limits to prove $$\lim_{n\to\infty}  \frac{n^3-n^2+9}{n^3-3n} =1 $$

So far I got 
$$ \left|  \frac{n^3-n^2+9}{n^3-3n} -1 \right| < \epsilon, $$
the left hand side is the same as
$$ \left| - \frac{n^2-3n-9}{n(n^2-3)} \right|$$
which we can then drop the absolute value and then make the inequality argument that
$$ \frac{n^2-3n-9}{n(n^2-3)} < \frac{n^2}{n(n^2-3)} =\frac{n}{(n^2-3)}.$$ 
From here I'm having trouble getting rid of the last $- 3$ to make the argument that $$\frac{1}{n} < \epsilon $$ which then means $\frac{1}{\epsilon} < n$
Should I choose a different $\epsilon$?  Is there some alternate way of expressing something that I'm missing. Thanks for the help!

Since you choose $n$ large anyway, you may assume $n^2 - 3 > \frac{1}{2} n^2$ (that is, you choose $n>\sqrt 2$, or simply $n>2$). Then 
$$\frac{n}{n^2 - 3}< \frac{n}{\frac 12 n^2} = \frac{2}{n}.$$
Now we want $\frac 2n<\epsilon$. So we need $n >\frac {2}{\epsilon}$. 
Combining the two information, one choose  
$$N = \max\{2, \frac{2}{\epsilon}\}. $$
