Question

limit X tends π/4 (sec^2x-2) / (tanx-1)

Answer 1

hope it helped u......

Answer 2

The value of $\lim_{x \to \frac{\pi}{4} } \dfrac{\sec^{2} x-2}{\tan x-1}$ is 2.

Step-by-step explanation:

We have,

$\lim_{x \to \frac{\pi}{4} } \dfrac{\sec^{2} x-2}{\tan x-1}$

To find, $\lim_{x \to \frac{\pi}{4} } \dfrac{\sec^{2} x-2}{\tan x-1}$ = ?

Put $x = \frac{\pi}{4}$, we get

$\dfrac{0}{0}$ form

=$\lim_{x \to \frac{\pi}{4} } \dfrac{1+\tan^{2} x-2}{\tan x-1}$

[ ∵$\sec^{2} A -\tan^{2} A=1$]

=$\lim_{x \to \frac{\pi}{4} } \dfrac{\tan^{2} x-1}{\tan x-1}$

=$\lim_{x \to \frac{\pi}{4} } \dfrac{(\tan x+1)(\tan x-1)}{\tan x-1}$

=$\lim_{x \to \frac{\pi}{4} } (\tan x+1)$

Put $x = \frac{\pi}{4}$, we get

$= \tan \dfrac{\pi}{4} +1$

= 1 + 1 = 2

Hence, the value of$\lim_{x \to \frac{\pi}{4} } \dfrac{\sec^{2} x-2}{\tan x-1}$ is 2.