Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 248: 69

$1$

Here, $\lim\limits_{x \to \dfrac{\pi}{2}^{-}} f(x)=\lim\limits_{x \to \dfrac{\pi}{2}^{-}} \dfrac{\sec x}{\tan x}$ This implies that $\lim\limits_{x \to \dfrac{\pi}{2}^{-}} (\dfrac{1}{\cos x})(\dfrac{\cos x}{\sin x})=\lim\limits_{x \to \dfrac{\pi}{2}^{-}} \dfrac{1}{\sin x}$ Thus, $\dfrac{1}{\sin \dfrac{\pi}{2}}=1$