Answer
Convergent to $2$.
Work Step by Step
$a_{n}=\frac{4n^{2}-3n}{2n^{2}+1}$
Dividing numerator and denominator by $n^{2}$, we have
$a_{n}=\frac{4-\frac{3}{n}}{2+\frac{1}{n^{2}}}$
$\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}\left(\frac{4-\frac{3}{n}}{2+\frac{1}{n^{2}}}\right)=\frac{4-0}{2+0}=2$
The sequence converges to $2$.
