1. ## Another Trigo-limit

Calculate this limit :
$
\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \left( {\sec x - \tan x} \right)
$

2. Originally Posted by dhiab
Calculate this limit :
$
\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \left( {\sec x - \tan x} \right)
$
There are numerous possible approaches including the use of l'Hopital's rule and power series. Please be clear on what approaches are NOT allowed.

3. sec x - tan x = (1/cos x) - (sin x)/(cos x) = (1 - sin x)/(cos x) =

(1 - sin x)/(cos x), rationalize the numerator by its conjugate, (1 + sin x).

= (1 - sin^2 x)/[(cos x)(1 + sin x],

= (1 - sin^2 x)/[(cos x)(1 + sin x)],

= (cos^2 x)/[(cos x)(1 + sin x)],

= [cos x cos x]/[cos x (1 + sin x)],

= ?

4. Hello, dhiab!

Calculate this limit: . $
\lim_{x\to\frac{\pi}{2}}(\sec x - \tan x)$

Multiply by $\frac{\sec x + \tan x}{\sec x + \tan x}$

. . $\frac{\sec x - \tan x}{1}\cdot\frac{\sec x + \tan x}{\sec x + \tan x} \;=\;\frac{\overbrace{\sec^2\!x - \tan^2\!x}^{\text{This is 1}}}{\sec x + \tan x} \;=\;\frac{1}{\sec x + \tan x}$

Then: . $\lim_{x\to\frac{\pi}{2}}\left(\frac{1}{\sec x + \tan x}\right) \;=\; \frac{1}{\sec\frac{\pi}{2} + \tan\frac{\pi}{2}} \;=\;\frac{1}{\infty + \infty} \;=\;0$