Chapter 1 - Limits and Continuity - 1.6 Continuity of Trigonometric Functions - Exercises Set 1.6 - Page 106: 26

$$1$$

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