Chapter 1 - Limits and Continuity - 1.6 Continuity of Trigonometric Functions - Exercises Set 1.6 - Page 106: 25

$$1$$

Given $$\lim_{t\to 0 }\frac{t^2}{1-\cos^2 t}$$ Then \begin{align*} \lim_{t\to 0 }\frac{t^2}{1-\cos^2 t}&=\lim_{t\to 0 }\frac{t^2}{\sin^2 t}\\ &=\lim_{t\to 0 }\frac{1}{\frac{\sin^2 t}{t^2}}\\ &=\frac{\lim_{t\to 0 }(1)}{\left(\lim_{t\to 0 }\frac{\sin t}{t }\right)^2}\\ &=\frac{1}{1}\\ &=1 \end{align*}