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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.

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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 at 16:21
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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 at 16:24
    
@SantiagoCanez yes, I know it. –  mattecapu May 21 at 16:26

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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 $n> \sqrt{\frac{2}{\varepsilon} - 1}$.

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For $\varepsilon \geq 2$ doesn't the inequality lose sense? $n$ would be complex –  mattecapu May 21 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 at 16:41
    
That's a perfectly good point. Thank you! Very helpful –  mattecapu May 21 at 16:47
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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 at 17:05

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