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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.

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    $\begingroup$ 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
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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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With your work:

$L=\lim\limits_{n\to\infty}\frac{n\sin n}{2n^2-1}=\lim\limits_{n\to\infty}\frac{\frac{\sin n}n}{\color{red}{\underbrace{2-\frac1{n^2}}_{\in [1,2)\ \forall n\geqslant 1}}}$

As mentioned: $$-1\leqslant\sin n\leqslant 1\Bigg/\cdot\frac1n\implies\boxed{-\underset{\Big\downarrow\\0}{\frac1n}\leqslant\frac{\sin n}n\leqslant\underset{\Big\downarrow\\0}{\frac1n}\implies \lim_{n\to\infty}\frac{\sin n}n=0}$$ $\implies L=\lim\limits_{n\to\infty}\frac{\frac{\sin n}n}{2-\frac1{n^2}}=0$

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