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I need to find $n_\epsilon$ such that for $\forall n \geq n_\epsilon : | x_n - \frac{1}{2} | < \epsilon$. So let's do that.

$$| \frac{n^2 + n + 1} {2\cdot(n^2 - 2)} - \frac{1}{2} | < \epsilon $$

$$| \frac{n + 3} {2\cdot(n^2 - 2)} | < \epsilon $$

That's the moment where I'm stucked, I don't know what to do next. Ok, maybe I can find some estimation (or evaluation I don't know how to say it properly) for epsilon in terms of n, so here it is:

$$-\epsilon < \frac{n + 3} {2\cdot(n^2 - 2)} < \epsilon $$ $$-\epsilon * 2n^2 - 4 < n + 3 < \epsilon * 2n^2 - 4$$ And it seems like dead end for me too, I can't figure out how estimate epsilon. How to proceed here? I have never come across situations like this when there is some "complex" fraction. Any tips are appreciated, thanks in advance!

RussianStranger
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  • Well, you could solve for $x$ the parametric system of inequations $$\begin{cases}x\ge 2&(\text{this condition was implictly used beforehand})\ -2\epsilon x^2-4<x+3\ x+3<2\epsilon x^2-4\end{cases}$$ and then observe that, for each $\epsilon$, the solution set $S_\epsilon$ contains a tailing set of $\Bbb N$. –  Oct 06 '21 at 11:00

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For $n > 4$, $$2(n^2-2) = 2n^2 -4 = n^2+n^2-4 > n^2 > 0$$ So $$|\frac{n+3}{2(n^2-2)}| < \frac{n+3}{n^2}<\frac{2n}{n^2}=\frac{2}{n}$$ hence take $n_{\epsilon} = \max\{4,[\frac{2}{\epsilon}]+1\}$.

zbo
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