Chapter 1 - Limits and Continuity - 1.6 Continuity of Trigonometric Functions - Exercises Set 1.6 - Page 105: 12

$$-\frac{\sqrt{3}}{2}$$

Given $$\lim_{x\to\infty}\sin \left(\frac{\pi x}{2-3x}\right) $$ Then \begin{align*} \lim_{x\to\infty}\sin \left(\frac{\pi x}{2-3x}\right) &=\sin \left(\lim_{x\to\infty}\frac{\pi x}{2-3x}\right)\\ &=\sin \left(\lim_{x\to\infty}\frac{\pi }{2/x-3}\right)\\ &=\sin \left( \frac{-\pi }{3}\right)\\ &=-\frac{\sqrt{3}}{2} \end{align*}