## Thomas' Calculus 13th Edition

$-1$
Let $\lim\limits_{x \to (\dfrac{\pi}{2})^{-}}f(x)=\lim\limits_{x \to (\dfrac{\pi}{2})^{-}} (x-\dfrac{\pi}{2}) (\sec x)$ But $f((\dfrac{\pi}{2})^{-})=\dfrac{0}{0}$ This shows that the limit has an indeterminate form, so we need to apply L-Hospital's rule as follows: $\lim\limits_{m \to n}f(x)=\lim\limits_{m \to n}\dfrac{p'(x)}{q'(x)}$ This implies that $\lim\limits_{x \to (\dfrac{\pi}{2})^{-}}\dfrac{1}{(-\sin x)}=-\dfrac{1}{\sin (\dfrac{\pi}{2})}=-1$