$\varepsilon$-$N$ 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!
 A: When you remove the modulus sign as well the negative sign you must ensure that the expression left is positive. Thus your first need is to ensure that $$\frac{n^{2}-3n-9}{n^{3}-3n}>0$$ The denominator is clearly positive if $n>3$ and the numerator is positive if $n>6$. Thus to proceed further we need to work with $n>6$. Next you can go a bit further using $$\frac{n} {n^{2}-3}<\frac{n}{n^{2}-3n}=\frac{1}{n-3}$$ You may further see that the expression is less than $2/n$ or you can work with the expression $1/(n-3)$ itself. Based on the approach you take the answer is $N=\max(6,2/\epsilon)$ or $N=\max(6,3+(1/\epsilon))$. Since these problems do not have a unique answer there is nothing to worry.

It is not clear why you want the final inequality to be $1/n<\epsilon $. It is not necessary that every limit problem will lead to that particular inequality. But in general one should strive to have an inequality of the form $a/(bn) <\epsilon$ under some assumption like $n>k$ and the answer is then $N=\max(k, a/(b\epsilon))$. In general one should not try to solve inequalities and obtain complicated expressions in terms of $\epsilon$ (most cheap textbooks are also at fault here). 
A: 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  
$$N = \max\{2, \frac{2}{\epsilon}\}. $$
