$\lim_{x\to \pi/2} \;\frac 1{\sec x+ \tan x}$ how to solve it answer is $0$, but $\frac 1{\infty + \infty}$ is indeterminate form
$$\lim_{x \to \pi/2} \frac 1{\sec x + \tan x}$$
 A: 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.
A: 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 
\[ \lim_{x \to \frac \pi 2} f(x) = \lim_{x\to\frac\pi 2} \frac{\cos x}{1 + \sin x} = \frac{\cos \frac \pi 2}{1 + \sin \frac \pi 2} = \frac{0}{1 + 1} = 0. \]
A: In order to solve this, it would be best to convert the $\sec x$ into $\frac{1}{\cos x}$. and convert $\tan x$ to $\frac{\sin x}{\cos x}$ So we will have
$$\lim_{x\to\frac{\pi}{2}} \frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}$$
Now we have to get rid of the fraction in the denominator. So multiply by $\cos x$ so we will have: 
$$\lim_{x\to\frac{\pi}{2}} \frac{1}{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}}$$
When then is turned into 
$$\lim_{x\to\frac{\pi}{2}} \frac{\cos x}{1 + \sin x}$$
Then sub $\pi/2$ and you'll be gold. 
$$\lim_{x\to\frac{\pi}{2}} \frac{\cos \frac{\pi}{2}}{1 + \sin \frac{\pi}{2}}$$
And the answer will be $0$. Therefore the limit is $0$.
A: Hint: "rationalize".
$$\frac 1{\sec x + \tan x}=\frac {\sec x - \tan x}{(\sec x)^2 - (\tan x)^2}=\sec x - \tan x=\frac{1-\sin(x)}{\cos(x)}=\frac{1-\sin^2(x)}{\cos(x)(1+\sin(x))}$$
A: $$\lim_{x \to \frac\pi2} \frac 1{\sec x + \tan x}$$
$$=\lim_{y \to 0} \frac 1{\sec \left(\frac\pi2-y\right) + \tan \left(\frac\pi2-y\right)}$$ (Putting $\frac\pi2-x=y,$ as $x \to \frac\pi2,y\to0$)
$$=\lim_{y \to 0} \frac 1{\csc y + \cot y}$$ 
$$=\lim_{y \to 0} \frac {\sin y}{1 + \cos y}$$ 
$$=\lim_{y \to 0} \frac {2\sin \frac y2\cos\frac y2}{2\cos^2\frac y2}$$ 
$$=\lim_{y \to 0} \tan  \frac y2$$ as $\cos\frac y2\ne0$
$$=0$$
