Question

The \displaystyle \lim_{x \to \frac{\pi}{2}} \left\\{ 2x \tan x - \frac{\pi}{\cos x }\right\\} is

• A. -1
• B. -3
• C. -2
• D. 0

Answer 1

Answer: (C)

-2

Answer 2

Let L =\displaystyle \lim _{x \rightarrow \frac{\pi}{2}}\left\\{2 x \tan x-\frac{\pi}{\cos X}\right\\}
=\displaystyle\lim _{x \rightarrow \frac{\pi}{2}}\left\\{\frac{2 x \sin x-\pi}{\cos \,x}\right\\}
=\displaystyle\lim _{x \rightarrow \frac{\pi}{2}}\left\\{2 x \frac{\sin x}{\cos X}-\frac{\pi}{\cos x}\right\\}
$[\frac{0}{0}$ form ]
Now, by using L'Hospital Rule
L= \displaystyle\lim _{x \rightarrow \frac{\pi}{2}}\left\\{\frac{2 \sin x+2 x \cos x}{(-\sin x)}\right\\}
$= \frac{2 \times 1+2 \times \frac{\pi}{2} \times \cos \frac{\pi}{2}}{(-1)}=-2$