Proving $ \lim\limits_{n\to\infty} \frac{8n^2-5}{4n^2+7} = 2$ using $\epsilon-\delta-$ definition.

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