Answer
$4$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to \frac{\pi}{2}^{-}} \dfrac{4 \tan x}{1+\sec x}$
Apply L-Hospital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
$ \lim\limits_{x \to \frac{\pi}{2}^{-}} \dfrac{4 \tan x}{1+\sec x}=\lim\limits_{x \to \frac{\pi}{2}^{-}} \dfrac{4 \sec^2 x}{\sec x \tan x}\\=\lim\limits_{x \to \frac{\pi}{2}^{-}} \dfrac{4 \sec x}{ \tan x}\\= 4 \lim\limits_{x \to \frac{\pi}{2}^{-}} cosec (x)\\=(4)(1)\\=4$
Thus, we have: $\lim\limits_{x \to \frac{\pi}{2}^{-}} \dfrac{4 \tan x}{1+\sec x}=4$
