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I was doing the proof to show that given $\epsilon > 0$, there exist an $n_ 0 \in \mathbb{N}$ such that etc.. And I chose my $n_0 > \max\lbrace2,\sqrt{(13+ \epsilon)/\epsilon} \rbrace$. But, is this correct?

I have that given an $\epsilon > 0$, choose $n_0 > \max\lbrace2,\sqrt{(13+ \epsilon)/\epsilon} \rbrace$. Then for all $n \ge n_0$, we have

$$\bigg|\frac{3n^2+5}{2n^2-1} - \frac{3}{2}\bigg| = \bigg| \frac{13}{2(2n^2-1)}\bigg| < \frac{13}{n^2-1}$$

but I am little confused for what to do next... What is the method?

And if not, how should I go about choosing my $n_0$ for situations similar to this?

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  • $\begingroup$ Note that if $n\geq n_0$, then $13/(n^2-1) <\epsilon$ $\endgroup$
    – user99914
    Oct 30, 2014 at 9:11
  • $\begingroup$ Thanks for the edits you guys, it is much more readable. Could you explain a little more? @John $\endgroup$
    – 9301293
    Oct 30, 2014 at 9:13
  • $\begingroup$ Note that $n > \sqrt{\frac{13+\epsilon}{\epsilon}}$ is the same as $\frac{13}{n^2-1} <\epsilon$. $\endgroup$
    – user99914
    Oct 30, 2014 at 9:20
  • $\begingroup$ so, if I choose $n_0>\sqrt{(14+\epsilon)/\epsilon}$, then, would this work? $\endgroup$
    – 9301293
    Oct 30, 2014 at 9:26
  • $\begingroup$ Yes, it works. Actually even $\sqrt{(13+\epsilon)/\epsilon}$ would do. $\endgroup$
    – user99914
    Oct 30, 2014 at 9:27

1 Answer 1

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Trusting your computations, you have $$ \bigg|\frac{3n^2+5}{2n^2-1} - \frac{3}{2}\bigg| = \bigg| \frac{13}{2(2n^2-1)}\bigg| \le \frac{13}{n^2} < \epsilon $$ with the last equality being true as soon as $$ n > n_\epsilon := \sqrt\frac{13}\epsilon $$ This is not the most accurate bound, but this is quick to find.

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