## Calculus 10th Edition

$\lim\limits_{x \to \infty} \frac{\sin x}{x-\pi} = 0$
As $\lim\limits_{x \to \infty} \frac{sinx}{x-\pi}$ the term $\sin x$ oscillates between 1 and -1. So the limit can be simplified into two different terms when infinity is substituted for x $-\frac{1}{\infty}$, and $\frac{1}{\infty}$, Both of which go to zero. So, it can be concluded that $\lim\limits_{x \to \infty} \frac{\sin x}{x-\pi} = 0$