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

We have for $x$ with $\cos x \ne 0$
\begin{align*}
  \frac 1{\tan x + \sec x} &= \frac 1{\frac{\sin x}{\cos x} + \frac 1{\cos x}}\\
   &= \frac{\cos x}{1 + \sin x}
\end{align*}
And hence 
\[ \lim_{x \to \frac \pi 2} f(x) = \lim_{x\to\frac\pi 2} \frac{\cos x}{1 + \sin x} = \frac{\cos \frac \pi 2}{1 + \sin \frac \pi 2} = \frac{0}{1 + 1} = 0. \]
