Use the $\epsilon-K$ definition to prove that the sequence $\frac{3n^2+2n+1}{4n^2+7}$ converges So far I know I want to use the triangle inequality and the fact that $\frac{1}{n^2} \leq \frac{1}{n}$ for all $n$ in the naturals. So my proof structure looks like this but then I get stuck.
Proof:
First, find the limit of the sequence. So $ \lim\limits_{n \to +\infty}\frac{3n^2+2n+1}{4n^2+7} = \frac{3}{4}$. So $x=\frac{3}{4}$.
Assume $\epsilon>0$.
So, $|\frac{3n^2+2n+1}{4n^2+7} - \frac{3}{4}| <\epsilon$.
= $|8n-\frac{17}{4n^2+7}|<\epsilon$.
=$|8n-\frac{17}{4n^2+7}| \leq |\frac{8n}{4n^2+7}| + |\frac{17}{4n^2+7}|$,by triangle inequality 
Then I was thinking to use the fact that $\frac{1}{n^2} < \frac{1}{n}$ and compare but I don't think my inequality is finished or clear. P.s Sorry I am new to this and trying to learn the syntax.
 A: We choose $\epsilon>0$ arbitrary and $n\in\mathbb{N}$ such that $n\ge \frac{1+\sqrt{1-28\epsilon^2-17\epsilon}}{4\epsilon}$, then you can check for yourself that $\lvert \frac{3n^2+2n+1}{4n^2+7}-\frac{3}{4}\rvert =\lvert \frac{8n-17}{4(4n^2+7)}\rvert<\epsilon$.
Hence $\forall\epsilon>0$$\exists n\in\mathbb{N}:\lvert\frac{3n^2+2n+1}{4n^2+7}-\frac{3}{4}\rvert<\epsilon.$ 
The key here is to solve the inequality $8n-17<\epsilon(4(4n^2+7))$. Actually you would have to consider different cases but since all terms with $n$ only have positive coefficients, you don't have to worry. Furthermore you can just solve the upper inequality as an equality, then pick the largest solution and choose your $n$ bigger than this solution, which is exactly what I did. This may not be the smallest $n\in\mathbb{N}$ to fulfill the desired condition, but it certainly works too.
A: The fraction should actually be $\frac{8n-17}{4(4n^2+7)}$ where the denominator looks like that. That is the aritmentic mistake. I don't know what next step to make from here.
