Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 have to verify the limit $$\lim_{n \to +\infty}\frac{2n^2}{n^2+1}=2$$ With $n\in\mathbb{N}$ (actually, it's the limit of an integer sequence).

So I set up the inequality $$\left| \, \frac{2n^2}{n^2+1} - 2 \, \right|\lt\varepsilon$$ And solving it, I get $$n^2\gt\frac{2}{\varepsilon} - 1$$ Now what? It isn't acceptable for $\varepsilon \geq 2$, thus what we conclude?

Intuitively I understand that the initial limit is true, but I can't come up with a formal justification.

share|cite|improve this question
In the formal definition of the limit of a sequence, if you have an $N$ which forms for one $\epsilon$, it automatically works for all larger $\epsilon$ as well. – Santiago Canez May 21 '14 at 16:21
It should be $n \to \infty$. If it is $x \to +\infty$ then the limit must be $\frac{2n^2}{n^2+1}$. – kmitov May 21 '14 at 16:24
@SantiagoCanez yes, I know it. – mattecapu May 21 '14 at 16:26
up vote 1 down vote accepted

To show convergence you must show that for any $\varepsilon>0$ there is an $N$ such that the inequality holds for all $n>N$. For $\varepsilon>2$ the inequality is automatically satisfied. For smaller $\epsilon$ you can take $n> \sqrt{\frac{2}{\varepsilon} - 1}$.

share|cite|improve this answer
For $\varepsilon \geq 2$ doesn't the inequality lose sense? $n$ would be complex – mattecapu May 21 '14 at 16:27
@mattecapu, I refer back to my comment above. The point is that you don't really have to worry about what happens for $\epsilon \ge 2$, since an $N$ which works for some $\epsilon < 2$ automatically works for any $\epsilon \ge 2$. – Santiago Canez May 21 '14 at 16:41
That's a perfectly good point. Thank you! Very helpful – mattecapu May 21 '14 at 16:47
Please, see what you have written. $\lim_{x \to +\infty}\frac{2n^2}{n^2+1}$. The fraction does not depend on $x$. – kmitov May 21 '14 at 17:05

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.