How large should n be for my choice of epsilon to work? I am having trouble finding the largest lower bound (inf?) on this sequence given my epsilon.
$$\frac{n^2 + 2n + 1}{2n^2 - n + 2} \to \frac{1}{2}$$
So $$\left|\frac{n^2 + 2n + 1}{2n^2 - n + 2}- \frac{1}{2}\right| = \left|\frac{5n}{4n^2 - 2n + 4}\right| \leq \frac{5}{2(n - 1)}$$
In the last fraction, I simply did $4n^2 -2n + 4 > 4n^2 - 2n = 2(2n^2 - n)$
Let's say I choose epsilon to be $10^{-6}$ (good enough I think)
My problem is that for this rational function, I could probably make an upper bound like $\frac{1}{n-1}$ and it would still work, but i am not sure how to keep doing this to get a "largest" lower bound
EDIT: I should probably almost add a remark that I would run into a similar problem for other rationals too
 A: The definition of limit says:
For all $\epsilon>0$ here is an $n_\epsilon$ ...
When you apply it you do not need to find the smallest $n_\epsilon$ that works; it is enough to find one. In any particular example it is enough that
$$
\frac{5}{2(n-1)}\le\epsilon\implies n\ge\frac{5}{2\epsilon}+1,
$$
so that you can take
$$
n_\epsilon=\left[\frac{5}{2\epsilon}+1\right]+1
$$
or any larger integer ($[z]$ is the integer part of $z$.) May be there is a smaller $n_\epsilon$ that works, but as long as the definition of limit is concerned, you don't care.
A: First a minor correction: the last displayed fraction should have $2(2n-1)$ in the denominator.
In proving that $f(n)\to L$ you don’t choose $\epsilon$: you show how to choose an $n_\epsilon$ (depending on $\epsilon$) such that $|f(n)-L|<\epsilon$ whenever $n\ge n_\epsilon$. You’ve shown for your $f(n)$ and $L=\frac12$ that 
$$|f(n)-L|\le\frac5{4n-2}$$
for all $n$. Thus, if you want to make $|f(n)-L|<\epsilon$, it’s good enough to make sure that $$\frac5{4n-2}<\epsilon$$ or, equivalently, that $$n>\frac{5+2\epsilon}{4\epsilon}=\frac5{4\epsilon}+\frac12\;.$$ Let $$n_\epsilon=\left\lceil\frac5{4\epsilon}\right\rceil+1\;,$$ where $\lceil x\rceil$ is the ceiling of $x$, the smallest integer $n$ such that $x\le n$; then for any $n\ge n_\epsilon$ it is true that $$|f(n)-L|\le\frac5{4n-2}<\epsilon\;.$$ We’ve given a recipe for constructing an $n_\epsilon$ big enough to do the job no matter what $\epsilon$ we’re given.
There’s point trying to be any more efficient; all that we need is some $n_\epsilon$ that does the job.
