How can I evaluate
$$ \lim_{n \rightarrow \infty} \frac{n\sin n}{2n^2- 1}? $$
Unsuccessful attempt:
In the expression $\frac{n\sin n}{2n^2 - 1}$, I divided the numerator and denominator by $n^2$, but I got stuck with $\frac{\sin n}{n}$ and I do not know how to go on.
Any help will be appreciated.
Since $-1\leq \sin x\leq 1$ for all $x$, we have $$ \left\lvert\frac{n\sin n}{2n^2-1}\right\rvert = \frac{n\lvert\sin n\rvert}{2n^2-1} \leq \frac{n}{2n^2-1}\,; $$ and since $2n^2-1\geq n^2$ for all $n\geq 1$, $$ \left\lvert\frac{n\sin n}{2n^2-1}\right\rvert \leq \frac{n}{n^2} = \frac{1}{n} \xrightarrow[n\to\infty]{}0. $$