# How can I evaluate $\lim_{n \rightarrow \infty} \frac{n\sin n}{2n^2 - 1}$?

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.

• when you multiply a sequence converging to zero by a bounded sequence you get a sequence converging to zero. Jun 11, 2019 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 $n\geq 1$, $$\left\lvert\frac{n\sin n}{2n^2-1}\right\rvert \leq \frac{n}{n^2} = \frac{1}{n} \xrightarrow[n\to\infty]{}0.$$
$$L=\lim\limits_{n\to\infty}\frac{n\sin n}{2n^2-1}=\lim\limits_{n\to\infty}\frac{\frac{\sin n}n}{\color{red}{\underbrace{2-\frac1{n^2}}_{\in [1,2)\ \forall n\geqslant 1}}}$$
As mentioned: $$-1\leqslant\sin n\leqslant 1\Bigg/\cdot\frac1n\implies\boxed{-\underset{\Big\downarrow\\0}{\frac1n}\leqslant\frac{\sin n}n\leqslant\underset{\Big\downarrow\\0}{\frac1n}\implies \lim_{n\to\infty}\frac{\sin n}n=0}$$ $$\implies L=\lim\limits_{n\to\infty}\frac{\frac{\sin n}n}{2-\frac1{n^2}}=0$$