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:
- 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 $N = \text{Max}\left(1, \dfrac{7}{9\epsilon}\right)$?