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