Proving $ \lim\limits_{n\to\infty} \frac{8n^2-5}{4n^2+7} = 2$ using $\epsilon-\delta-$ definition. 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}} $
$$ \left|\frac{8n^2-5}{4n^2+7} -2 \right| \leq \frac{19}{4n^2} < \frac{19}{4(\frac{19}{4\epsilon})} = \epsilon $$

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