Limits with sine and a fraction

Question

$\lim\limits_{x\to\infty}\dfrac{2}{x^2-1}{sinx}=?$

Personal work :

For sure, $\lim\limits_{x\to\infty}\dfrac{2}{x^2-1}=\lim\limits_{x\to\infty}\dfrac{2}{x^2}=0^+$

But what about $\lim\limits_{x\to\infty}sinx?$

Answer 1

By squeeze theorem

$$0\le\left|\frac{2}{x^2-1}{\sin x}\right|<\frac{2}{x^2-1}\to 0$$

Answer 2

The limit is zero because $sin$ is bounded. $\lim\limits_{x\rightarrow a} fg=0$ if $\lim\limits_{x\rightarrow a}f=0$ and $g$ is bounded.

Answer 3

\begin{align*} \left|\dfrac{2\sin x}{x^{2}-1}\right|\leq\dfrac{2}{x^{2}-x^{2}/2}=\dfrac{4}{x^{2}}\leq\dfrac{4}{x}\rightarrow 0 \end{align*} whenever $x\rightarrow\infty$.