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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?

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    Ok, I give up. I$n$ Italy we used to do these things in high school (although the level is now lower than it was in the 90s), and the general feeling is that real analysis is the subject covered by Royden's or Rudin's books.2012-09-13

2 Answers 2

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You want to prove that $\left|\frac{-n-7}{n^2-2n} \right| < \epsilon$ provided that $n>K(\epsilon)$. You can write $ \left|\frac{-n-7}{n^2-2n} \right| \leq \left| \frac{2n}{\frac{1}{2}n^2}\right| = \frac{4}{n}, $ at least for $n \gg 1$.

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    Yes. And you can also find explicitly the threshold beyond which these estimates are true. You can then take $K(\epsilon)$ to be the largest threshold.2012-09-12
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The limit is much easier calculated by observing that $\frac {2n^2-3n-5} {n^2-2n+2} = \frac {2 - {3 \over n} - {5 \over n^2}}{1 - {2 \over n} + {2 \over n^2}}$

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    Jesse is right. I was trying to do so with the definition. However, I see I should have clarified that. I basically just found a K(e$p$silon) that existe$d$ that satisfied it.2012-09-19