$\lim _{x \rightarrow \dfrac{\pi}{4}} \dfrac{\sec ^2 x-2}{\tan x-1}=$

Given the limit $L = \lim _{x \rightarrow \dfrac{\pi}{4}} \dfrac{\sec ^2 x-2}{\tan x-1}$.

Using the identity $\sec ^2 x = 1 + \tan ^2 x$, the expression becomes:

$L = \lim _{x \rightarrow \dfrac{\pi}{4}} \dfrac{1 + \tan ^2 x - 2}{\tan x - 1}$

$L = \lim _{x \rightarrow \dfrac{\pi}{4}} \dfrac{\tan ^2 x - 1}{\tan x - 1}$

Factoring the numerator as a difference of squares:

$L = \lim _{x \rightarrow \dfrac{\pi}{4}} \dfrac{(\tan x - 1)(\tan x + 1)}{\tan x - 1}$

Canceling the common term $(\tan x - 1)$ for $x \neq \dfrac{\pi}{4}$:

$L = \lim _{x \rightarrow \dfrac{\pi}{4}} (\tan x + 1)$

Substituting $x = \dfrac{\pi}{4}$:

$L = \tan \left(\dfrac{\pi}{4}\right) + 1 = 1 + 1 = 2$.

Answer: 2