Answer
Sequence converges to $\frac{1}{2}$
Work Step by Step
$a_{n}=\sqrt (\frac{3+2n^{2}}{8n^{2}+n})$
$\lim\limits_{n \to \infty}\sqrt (\frac{3+2n^{2}}{8n^{2}+n})$
$\lim\limits_{n \to \infty}\sqrt (\frac{\frac{3}{n^{2}}+2}{8+\frac{1}{n}})$
$\lim\limits_{n \to \infty} \sqrt (\frac{0+2}{8+0})$
$\lim\limits_{n \to \infty} \sqrt (\frac{1}{4})$
Sequence converges to $\frac{1}{2}$
