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

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 $0$.
