Question

$L=\underset{x\to \pi /2}{lim}[x\tan x-\left(\pi /2\right)\sec x]$
Then value of $L+2$ is

Answer 1

Let, $L=\underset{x\to \pi /2}{lim}[x\tan x-\left(\pi /2\right)\sec x]$
$=\underset{x\to \pi /2}{lim}\frac{2x\sin x-\pi }{2\cos x}\left[Form\frac{0}{0}\right]$
Thus using L-Hospital's rule,
$L==\underset{x\to \pi /2}{lim}\frac{(2\sin x+2x\cos x)}{-2\sin x}$
$=\frac{2\times 1+2\times \pi /2\times 0}{-2.1}=-1.$

$L+2=1$