Answer
$\frac{3}{2}$
Work Step by Step
By dividing the numerator and the denominator by $n^2$, we have
$$ \lim\limits_{n \to \infty}{a_n}=\lim\limits_{n \to \infty}\frac{3n^2+n+2}{2n^2-3}=\frac{3}{2} .$$
Hence, using Theorem 1, we see that the sequence $a_n$ converges to $\frac{3}{2}$.
