Question

$\underset{x\to \frac{\pi }{2}}{lim}\frac{1-(\sin x{)}^{\sin x}}{{\cos }^{2}x}=$

• A. $2$
• B. $1$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $\frac{1}{2}$

$\underset{x\to \frac{\pi }{2}}{lim}\frac{1-(\sin x{)}^{\sin x}}{{\cos }^{2}x}ApplyingL-Hospita{l}^{\prime }srule,=\underset{x\to \frac{\pi }{2}}{lim}\frac{-(\sin x{)}^{\sin x}\cos x(\log \sin x+1)}{-2\sin x\cos x}=\underset{x\to \frac{\pi }{2}}{lim}\frac{(\sin x{)}^{\sin x}(\log \sin x+1)}{2\sin x}=\frac{1}{2}$
Hence, option 'C' is correct.