Answer
We prove using the limit definition:
$\mathop {\lim }\limits_{n \to \infty } \dfrac{{2n - 1}}{{3n + 2}} = \dfrac{2}{3}$
Work Step by Step
Let $f\left( x \right) = \dfrac{{2n - 1}}{{3n + 2}}$.
To prove that $\mathop {\lim }\limits_{n \to \infty } \dfrac{{2n - 1}}{{3n + 2}} = \dfrac{2}{3}$, we must show that $\left| {f\left( x \right) - \dfrac{2}{3}} \right|$ becomes arbitrarily small when $n$ approaches $\infty $.
We have
$\left| {f\left( x \right) - \dfrac{2}{3}} \right| = \left| {\dfrac{{2n - 1}}{{3n + 2}} - \dfrac{2}{3}} \right| = \left| {\dfrac{{6n - 3 - 6n - 4}}{{3\left( {3n + 2} \right)}}} \right| = \left| {\dfrac{{ - 7}}{{3\left( {3n + 2} \right)}}} \right|$
As $n \to \infty $, we have $\left| {f\left( x \right) - \dfrac{2}{3}} \right| \to 0$ as is required. Therefore, by definition $\mathop {\lim }\limits_{n \to \infty } \dfrac{{2n - 1}}{{3n + 2}} = \dfrac{2}{3}$.
