$\epsilon-N$ for $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$. I want to prove using $\epsilon-N$ that $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$.
Firstly we will find a sufficently large $n\in \Bbb N$:
\begin{align*}
\epsilon\quad\gt&\quad\left|\frac{3n+1}{2n+5} - \frac32 \right|\\
=&\quad \left|\frac{3n+1 -3n-7.5}{2n+5}\right|\\
=&\quad \left|\frac{-13}{4n+10}\right|\\
\implies&\quad \epsilon \gt \frac{13}{4n+10}\\
\implies&\quad \frac1\epsilon \lt \frac{4n+10}{13}\\
\implies&\quad \frac14\left(\frac{13}\epsilon-10\right) \lt n\\
\end{align*}
So $\forall \epsilon\gt 0, \exists N(\epsilon)>\frac14\left(\frac{13}\epsilon-10\right)$ such that $\forall n\in N, n>N$, $a_n=|\frac{3n+1}{2n+5}-\frac32|\lt \epsilon$
Proof: $\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right| \color{red}{\lt} \left|\frac{3N+1}{2N+5}-\frac{3}{2}\right|\lt\epsilon$
Assuming this is even right. Everything feels good except the $\color{red}{\lt}$ in red, and then when I check, since we have 
\begin{align*}
&\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right| \color{red}{\lt}\left|\frac{3N+1}{2N+5}-\frac{3}{2}\right|,\quad n\gt N\\
&\implies \left|\frac{3n+1}{2n+5} - \frac32\right| - \left|\frac{3(n+1)+1}{2(n+1)+5} - \frac{3}{2}\right|\gt 0
\end{align*}

But this fails for all $n$? What did I do wrong?
Also, since $N\in\Bbb N$, we have to choose our $N$ so it is a natural number simultaneous to being greater than $\frac14\left(\frac{13}\epsilon-10\right)$, how do I ensure that subbing in $\frac14\left(\frac{13}\epsilon-10\right)$ is greater than subbing in the natural number that is the ceiling of this?


$\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac32$.
Let $\epsilon >0$ be arbitrary.
Now by the archemedian property there exists $N \in \mathbb N$ such that $N>\frac{13}{\epsilon}$.
Observe that for each $n>N $
$|\frac{3n+1}{2n+5}-\frac{3}{2}|=\frac{13}{4n+10}< \frac{13}{n}<\frac{13}{N}<\epsilon$.
