I wish to prove that a certain rational function has a certain limit.
The question is: what is $\lim_{n\to\infty} \frac{2n^2-3n-5}{n^2-2n+2}?$
Obviously the limit is $2$.
My attempt to prove it:
Suppose $\epsilon > 0$. Then let $K(\epsilon)$ be an element of $\mathbb N$ such that $K(\epsilon) > \text{??}$
I know that I have to start with this: $\left|\frac{2n^2-3n-5}{n^2-2n+2} - 2\right|$
Subtracting, I get $\left|\frac{-n-7}{n^2-2n+2}\right|$ I can reduce this by saying that the last term is less than $\left|\frac{-n-7}{n^2-2n}\right|$.
I know that I need to set this equal to $\epsilon$ but I think there must be a way to further reduce this. Any help?