## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.1 Sequences - Exercises - Page 538: 43

#### 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 $\frac{3}{2}$.

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