$\lim_{x\to \pi/2} \;\frac 1{\sec x+ \tan x}$ how to solve it answer is $0$, but $\frac 1{\infty + \infty}$ is indeterminate form
$$\lim_{x \to \pi/2} \frac 1{\sec x + \tan x}$$

Clarification:
$$\lim_{x \to \left(\frac{\pi}{2}\right)^+} \frac{1}{\sec x + \tan x} \to \frac{1}{\infty} = 0$$
$$\lim_{x \to \left(\frac{\pi}{2}\right)^-} \frac{1}{\sec x + \tan x} \to -\frac{1}{\infty} = 0$$
$$\lim_{x \to \frac{\pi}{2}} \frac 1{\sec x + \tan x} \to \frac{1}{\infty} = 0$$
In other words, the limit, $\,\to \frac 1{\infty}\,$ is not of indeterminate form: the limit in both cases above is equal to zero.
I do believe that recognizing your function $$f(x) = \frac 1{\tan x + \sec x} = \frac{\cos x}{1 + \sin x}$$ makes the limit as $x \to \pi/2$ perhaps more evident.
