Proof of an Limit Using the formal definition of convergence, Prove that $\lim\limits_{n \to \infty} \frac{3n^2+5n}{4n^2 +2} = \frac{3}{4}$.
Workings:
If $n$ is large enough, $3n^2 + 5n$ behaves like $3n^2$
If $n$ is large enough $4n^2  + 2$ behaves like $4n^2$
More formally we can find $a,b$ such that $\frac{3n^2+5n}{4n^2 +2} \leq \frac{a}{b} \frac{3n^2}{4n^2}$
For $n\geq 2$ we have $3n^2 + 5n /leq 3n^2.
For $n \geq 0$ we have $4n^2 + 2 \geq \frac{1}{2}4n^2$
So for $ n \geq \max\{0,2\} = 2$ we have:
$\frac{3n^2+5n}{4n^2 +2} \leq \frac{2 \dot 3n^2}{\frac{1}{2}4n^2} = \frac{3}{4}$ 
To make $\frac{3}{4}$ less than $\epsilon$:
$\frac{3}{4} < \epsilon$, $\frac{3}{\epsilon} < 4$
Take $N = \frac{3}{\epsilon}$ 
Proof:
Suppose that $\epsilon > 0$
Let $N = \max\{2,\frac{3}{\epsilon}\}$
For any $n \geq N$, we have that $n > \frac{3}{\epsilon}$ and $n>2$, therefore
$3n^2 + 5n^2 \leq 6n^2$ and $4n^2 + 2 \geq 2n^2$
Then for any $n \geq N$ we have
$|s_n - L| = \left|\frac{3n^2 + 5n}{4n^2 + 2} - \frac{3}{4}\right|$
$ = \frac{3n^2 + 5n}{4n^2 + 2} - \frac{3}{4}$
$ = \frac{10n-3}{8n^2+4}$
Now I'm not sure on what to do. Any help will be appreciated.

Perhaps simpler:
With the Squeeze Theorem:
$$\frac34\xleftarrow[x\to\infty]{}\frac{3n^2}{4n^2}\le\frac{3n^2+5n}{4n^2+2}\le\frac{3n^2+5n}{4n^2}=\frac34+\frac54\frac1{n}\xrightarrow[n\to\infty]{}\frac34+0=\frac34$$
With arithmetic of limits:
$$\frac{3n^2+5n}{4n^2+2}=\frac{3n^2+5n}{4n^2+2}\cdot\frac{\frac1{n^2}}{\frac1{n^2}}=\frac{3+\frac5n}{4+\frac2{n^2}}\xrightarrow[n\to\infty]{}\frac{3+0}{4+0}=\frac34$$
