Answer
$1$
Work Step by Step
$\lim\limits_{x \to \dfrac{\pi}{2}^{-}} f(x)=\lim\limits_{x \to \dfrac{\pi}{2}^{-}} \dfrac{\sec x}{\tan x}$
The given function can be re-arranged as:
$\lim\limits_{x \to \dfrac{\pi}{2}^{-}} (\dfrac{1}{\cos x})(\dfrac{\cos x}{\sin x})=\lim\limits_{x \to \dfrac{\pi}{2}^{-}} \dfrac{1}{\sin x}$
Hence, $\dfrac{1}{\sin \dfrac{\pi}{2}}=\dfrac{1}{1}=1$
