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!
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).