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.

  • $\begingroup$ when you multiply a sequence converging to zero by a bounded sequence you get a sequence converging to zero. $\endgroup$ – PierreCarre Jun 11 '19 at 10:21

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 $n\geq 1$, $$ \left\lvert\frac{n\sin n}{2n^2-1}\right\rvert \leq \frac{n}{n^2} = \frac{1}{n} \xrightarrow[n\to\infty]{}0. $$


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