Answer
converges to $\dfrac{1}{2}$
Work Step by Step
Let $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{n^2 \sin (1/n)}{2n-1}$
But $ \lim\limits_{n \to \infty} \dfrac{n^2 \sin (1/n)}{2n-1}=\dfrac{0}{0}$
This shows that the limit has an Indeterminate form, so we will use L-Hospital's rule.
This implies that $ \lim\limits_{n \to \infty} \dfrac{-\cos (1/n) \cdot (1/n^2)}{(-2n/n^2)+(2/n^3)}=\lim\limits_{n \to \infty} \dfrac{-\cos (1/n)}{-2+(2/n)}$
and, $\lim\limits_{n \to \infty} \dfrac{-\cos (1/n)}{-2+(2/n)}=\dfrac{1}{2}$
Thus, $\lim\limits_{n \to \infty} a_n=\dfrac{1}{2}$ and {$a_n$} converges to $\dfrac{1}{2}$.
