Prove $\lim\limits_{n \to\infty} a_n =\lim\limits_{n \to\infty} \frac{2n-1}{3n+2} = \frac{2}{3}$ Prove: $\lim\limits_{n \to\infty} a_n =\lim\limits_{n \to\infty} \dfrac{2n-1}{3n+2} = \dfrac{2}{3}$ using the definition of the limit.
This is what I have so far:


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*Let $\epsilon > 0$ and take $N = \text{Max}\left(1, \dfrac{7}{9\epsilon}\right)$. This is my reasoning:


Solve: $\left|\dfrac{2n-1}{3n+2} - \dfrac{2}{3}\right| < \epsilon$
We get: $\left|\dfrac{-7}{9n+6}\right| < \epsilon$ $\iff$ $\left|\dfrac{-7}{9n+6}\right| < \left|\dfrac{7}{9n}\right| < \epsilon$
Take $n > 1$ so we can drop the absolute value sign and Solve for $n$:
$\dfrac{7}{9\epsilon} < n$
So where would I go from here? Also, does taking $n > 1$ mean we have to do $N = \text{Max}\left(1, \dfrac{7}{9\epsilon}\right)$?

$\left|\dfrac{2n-3}{3n+2} - \dfrac{2}{3}\right| < \epsilon$
$\implies\left|\dfrac{-13}{9n+6}\right| < \epsilon$ $ \text{ and }  $$\left|\dfrac{-13}{9n+6}\right| < \left|\dfrac{13}{9n}\right| < \epsilon$
Thus, 
Given $\epsilon\gt 0,$
For $\forall n\gt \dfrac{13}{9\epsilon},  \left|\dfrac{2n-3}{3n+2} - \dfrac{2}{3}\right| < \epsilon $
Hence the limit of $\dfrac{2n-3}{3n+2} $ is $\dfrac{2}{3}$
