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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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    I think that the label real-analysis is really too much for this question.2012-09-12
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    I disagree. This is the kind of question that students deal with in a first course on real analysis. At least in the US, the regular three-semester calculus sequence typically does not cover delta-epsilon proofs.2012-09-12
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    @Jesse: I am under the impression that, at least in the US, epsilon-delta proofs are the very first thing taught: just not as extensively as one would need for a real analysis course.2012-09-12
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    @Hurkyl: My understanding is that in most calculus classes, the actual epsilon-delta definition is mentioned, but students aren't ever required to use it.2012-09-12
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    Ok, I give up. In 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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    can you do that because 2n is much greater than abs(-n-7) and 1/2n^2 is much smaller than n^2 - 2n for large values of n2012-09-12
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    For every $n\geqslant7$.2012-09-12
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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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    The OP isn't asking how to calculate the value of the limit, but rather how to _prove_ that the limit is 2.2012-09-12
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    By calculating the limit one also *proves* that the limit is 2.2012-09-12
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    Let me rephrase. I mean that the OP is looking for a $K(\epsilon)$ such that $n > K(\epsilon)$ implies $$\left|\frac{-n-7}{n^2-2n}\right| < \epsilon.$$ Maybe I'm missing something, but I don't see how your hint helps find that $K(\epsilon)$ or complete the proof that the OP has started.2012-09-12
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    I fully agree that if one has proven that $\lim 1/n = \lim 1/n^2 = 0$, and if one has also proven the rules for limits of sums and quotients, then yes, your solution provides a proof. My point is simply that such a proof is not in the spirit of the question.2012-09-12
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    I see your point, but I would like to stress that proving something from definition is not always desirable. And unless there are good reasons (which OP does not mention) to dismiss the proof using the calculation over the proof from definition, the former is preferable.2012-09-13
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    OK, Fair enough.2012-09-13
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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(epsilon) that existed that satisfied it.2012-09-19