Answer
$\lim\limits_{x \to (\pi/2)^+}e^{\sec x}=0$
Work Step by Step
$\lim\limits_{x \to (\pi/2)^+}e^{\sec x}$ (By letting $y=\sec x$, as $x\to (\pi/2)^+$, $y\to -\infty$)
$=\lim\limits_{y \to -\infty}e^y$ (By letting $t=-y$, as $y\to -\infty $, $t\to \infty$)
$=\lim\limits_{t \to \infty}e^{-t}$
$=\lim\limits_{t \to \infty}\frac{1}{e^t}$ (By letting $w=e^t$, as $t\to \infty $, $w\to \infty$)
$=\lim\limits_{w \to \infty}\frac{1}{w}$
$=0$
Thus, $\lim\limits_{x \to (\pi/2)^+}e^{\sec x}=0$.
