Question

Solve : $\underset{x⟶\frac{\pi }{2}}{lim}\frac{{1-\left(sinx\right)}^{sinx}}{{cos}^{2}x}=$

• A. 2
• B. 1
• C. 1/2
• D. 1/4

Answer 1

The correct option is C 1/2

$L=\underset{x⟶\frac{\pi }{2}}{lim}\frac{{1-\left(sinx\right)}^{sinx}}{{cos}^{2}x}$

Given limit is of form $\frac{0}{0}$. So, using $L^{\prime }Hopita{l}^{\prime }s$ rule:

$=\underset{x⟶\frac{\pi }{2}}{lim}\frac{{0-\frac{d}{dx}\left(sinx\right)}^{sinx}}{-2cosx\sin x}$

$=\underset{x⟶\frac{\pi }{2}}{lim}\frac{{\left(sinx\right)}^{sinx}[\cos x\ln \sin x+\cos x]}{\sin 2x}=\frac{0}{0}$

Again it is $\frac{0}{0}$ form. So differentiating again we get:

$=\underset{x⟶\frac{\pi }{2}}{lim}\frac{{\left(sinx\right)}^{sinx}[\cos x\frac{1}{\sin x}\cos x-\sin x\ln \sin x-\sin x]+\sin x^{\sin x}[\cos x\ln \sin x\cos x{]}^2}{2\cos 2x}$

$=\frac{1[0-0-1]+1[0+0]}{-2}$

$=\frac{1}{2}$