The problem asks us to find the limit of the sequence $a_n = \frac{\sqrt{3n^2 + 2}}{2n + 1}$ as $n$ approaches infinity.

1. Problem Description

The problem asks us to find the limit of the sequence $a_n = \frac{\sqrt{3n^2 + 2}}{2n + 1}$ as $n$ approaches infinity.

2. Solution Steps

To find the limit of the sequence as $n$ approaches infinity, we can divide both the numerator and the denominator by the highest power of $n$ that appears in the expression. In this case, it's $n$. We need to be careful with the numerator since it contains a square root.

$a_n = \frac{\sqrt{3n^2 + 2}}{2n + 1}$

We divide the numerator and denominator by $n$. In the numerator, we divide by $n = \sqrt{n^2}$.

$a_n = \frac{\frac{\sqrt{3n^2 + 2}}{n}}{\frac{2n + 1}{n}} = \frac{\frac{\sqrt{3n^2 + 2}}{\sqrt{n^2}}}{\frac{2n}{n} + \frac{1}{n}} = \frac{\sqrt{\frac{3n^2 + 2}{n^2}}}{2 + \frac{1}{n}} = \frac{\sqrt{3 + \frac{2}{n^2}}}{2 + \frac{1}{n}}$

Now, we take the limit as $n$ approaches infinity:

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\sqrt{3 + \frac{2}{n^2}}}{2 + \frac{1}{n}}$

As $n \to \infty$, $\frac{2}{n^2} \to 0$ and $\frac{1}{n} \to 0$. Therefore,

$\lim_{n \to \infty} a_n = \frac{\sqrt{3 + 0}}{2 + 0} = \frac{\sqrt{3}}{2}$

3. Final Answer

$\frac{\sqrt{3}}{2}$