CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

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


Solution

Let, $$L=\displaystyle \lim_{x\rightarrow \pi /2}\left [ x\tan x-\left ( \pi /2 \right )\sec x \right ]$$
$$\displaystyle = \lim_{x\rightarrow \pi /2}\frac{2x\sin x-\pi }{2\cos x} \left [ Form\frac{0}{0} \right ]$$
Thus using L-Hospital's rule,
$$L=\displaystyle 
= \lim_{x\rightarrow \pi /2}\frac{\left ( 2\sin x+2x\cos x \right )}{-2\sin x}$$
$$\displaystyle = \frac{2\times1+2\times\pi /2\times0}{-2.1}= -1.$$
$$L+2=1$$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image