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)

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\} $
$= \frac{2 \times 1+2 \times \frac{\pi}{2} \times \cos \frac{\pi}{2}}{(-1)}=-2$