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!
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 $n_{\epsilon} = \max\{4,[\frac{2}{\epsilon}]+1\}$.