Question

$\underset{x\to \frac{\pi }{2}}{\lim }\frac{\sin x-(\sin x{)}^{\sin x}}{1-\sin x+\ln \sin x}$ equals

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

Let $\sin x=h$, then as $x\to \pi /2,h\to 1$
$\therefore$ given limit
$=\underset{h\to 1}{\lim }\frac{h-h^h}{1-h+\ln h}$
$=\underset{h\to 1}{\lim }\frac{1-h^h-h^h\ln h}{-1+1/h}$
(Using L’ Hospital Rule)
$=\underset{h\to 1}{\lim }\frac{-h^h-2h^h\ln h-h^{h-1}-h^h(\ln h{)}^2}{-1/h^2}$
$=\frac{-1-1}{-1}=2$