Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
I think that the label real-analysis is really too much for this question. – Siminore Sep 12 '12 at 17:28
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. – Jesse Madnick Sep 12 '12 at 19:05
@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. – Hurkyl Sep 12 '12 at 19:11
@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. – Jesse Madnick Sep 12 '12 at 19:17
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. – Siminore Sep 13 '12 at 7:48
up vote 1 down vote accepted

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 $n \gg 1$.

share|cite|improve this answer
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 n – Samuel Gregory Sep 12 '12 at 17:29
For every $n\geqslant7$. – Did Sep 12 '12 at 17:30
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. – Siminore Sep 12 '12 at 17:31

The limit is much easier calculated by observing that $$\frac {2n^2-3n-5} {n^2-2n+2} = \frac {2 - {3 \over n} - {5 \over n^2}}{1 - {2 \over n} + {2 \over n^2}}$$

share|cite|improve this answer
The OP isn't asking how to calculate the value of the limit, but rather how to prove that the limit is 2. – Jesse Madnick Sep 12 '12 at 19:09
By calculating the limit one also proves that the limit is 2. – Levon Haykazyan Sep 12 '12 at 20:05
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. – Jesse Madnick Sep 12 '12 at 21:56
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. – Jesse Madnick Sep 12 '12 at 22:00
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. – Levon Haykazyan Sep 13 '12 at 10:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.