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I'm trying to prove the limit of this sequence using the formal definition. I've looked at other questions on the site but from the ones I've seen, the $n^2$ term always seems to cancel out, making it simpler.

Show that $$ \lim_{n\to\infty} \frac{8n^2-5}{4n^2+7} = 2$$

So this is how I started:

$$ \left|\frac{8n^2-5}{4n^2+7} -2 \right| = \left|\frac{-19}{4n^2+7}\right| = \frac{19}{4n^2+7} \leq \frac{19}{4n^2} = \epsilon$$

Let $\epsilon > 0 \implies n = \sqrt{\frac{19}{4\epsilon}}$

By Archimedian Property: $ N > \sqrt{\frac{19}{4\epsilon}} $

If $ n \geq N \geq \sqrt{\frac{19}{4\epsilon}} $

$$ \left|\frac{8n^2-5}{4n^2+7} -2 \right| \leq \frac{19}{4n^2} < \frac{19}{4(\frac{19}{4\epsilon})} = \epsilon $$

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  • $\begingroup$ Looks correct to me. $\endgroup$ – user491874 Nov 18 '17 at 21:23
  • $\begingroup$ Welcome to math.stackexhange! This looks fine. $\endgroup$ – Hans Engler Nov 18 '17 at 21:24
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    $\begingroup$ +1 for using MathJax and showing your attempted solution. $\endgroup$ – Remy Nov 18 '17 at 21:26
  • $\begingroup$ Yes, it’s correct. Just one small point, use “there exist” before N in the step where you use Archimedian property. $\endgroup$ – Mayuresh L Nov 18 '17 at 21:28
  • $\begingroup$ Canceling out the $n^2$ will leave $\frac 5{n^2}$ and $\frac 7{n^2}$. I wouldn't say that is any easier. Maybe more intuitive where to go, but not easier.... any way, what you did was exactly what you were supposed to do and you did it exactly right. (Ithink... $\frac {8n^2 - 5}{4n^2 + 7} - 2 = \frac {(8n^2 - 5) - (8n^2 + 14)}{4n^2 + 7} = \frac {-19}{4n^2 + 7} ....$ Yep, it checks out. (Lack of an $n$ term threw me but it's fine...) $\endgroup$ – fleablood Nov 18 '17 at 22:35
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we have $$\left|\frac{8n^2-5}{4n^2+7} -2 \right| = \left|\frac{-19}{4n^2+7}\right| < \frac{19}{4n^2} = $$

Now let such that $$\frac{19}{4n^2} < \epsilon\Longleftrightarrow n>\sqrt{\frac{19}{4\epsilon}} $$ Now choosing,

$$N =\left\lfloor \sqrt{\frac{19}{4\epsilon}}\right\rfloor +1$$ then we have $$n > N=\left\lfloor \sqrt{\frac{19}{4\epsilon}}\right\rfloor +1\Longrightarrow n>\sqrt{\frac{19}{4\epsilon}}\\\Longrightarrow \epsilon>\frac{19}{4n^2} > \left|\frac{8n^2-5}{4n^2+7} -2 \right|$$

that is we have $$n > N=\left\lfloor \sqrt{\frac{19}{4\epsilon}}\right\rfloor +1\Longrightarrow \left|\frac{8n^2-5}{4n^2+7} -2 \right|< \epsilon$$

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