Chapter 1 - Limits and Continuity - 1.6 Continuity of Trigonometric Functions - Exercises Set 1.6 - Page 106: 24

$$\infty$$

Given $$\lim_{h\to 0 }\frac{\sin h}{1-\cos h}$$ Then \begin{align*} \lim_{h\to 0 }\frac{\sin h}{1-\cos h}&=\lim_{h\to 0 }\frac{\frac{\sin h}{h}}{\frac{1-\cos h}{h}}\\ &=\frac{\lim_{h\to 0 }\frac{\sin h}{h}}{\lim_{h\to 0 }\frac{1-\cos h}{h}} \\ &=\frac{1}{0}\\ &=\infty \end{align*}