Answer
$\lim\limits_{x \to (\pi/2)^-}\frac{sec~x}{tan~x} = 1$
Work Step by Step
$\lim\limits_{x \to (\pi/2)^-}\frac{sec~x}{tan~x} = \frac{\infty}{\infty}$
We can try to apply L'Hospital's Rule:
$\lim\limits_{x \to (\pi/2)^-}\frac{sec~x~tan~x}{sec^2~x} = \lim\limits_{x \to (\pi/2)^-}\frac{tan~x}{sec~x} = \frac{\infty}{\infty}$
Applying L'Hospital's Rule does not help us solve the question.
We can try another method:
$\lim\limits_{x \to (\pi/2)^-}\frac{sec~x}{tan~x}$
$= \lim\limits_{x \to (\pi/2)^-}\frac{1/cos~x}{sin~x/cos~x}$
$= \lim\limits_{x \to (\pi/2)^-}\frac{cos~x/cos~x}{sin~x}$
$= \lim\limits_{x \to (\pi/2)^-}\frac{1}{sin~x}$
$= \frac{1}{1}$
$= 1$
Therefore:
$\lim\limits_{x \to (\pi/2)^-}\frac{sec~x}{tan~x} = 1$