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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

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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.

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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.