`underset(xto 1^-)lim(sqrtpi-sqrt(2sin^-1x))/(sqrt(1-x))` is equal to

Options

A. `sqrt(pi)`
B. `sqrt((2)/(pi))`
C. `(1)/(sqrt(2pi))`
D. `sqrt((pi)/(2))`

Answer

B

Explanation

`lim_(x rarr 1^(-))(sqrt(pi)-sqrt(2sin^(-1)x))/(sqrt(1-x)) = lim_(x rarr 1^(-))(pi - 2 sin^(-1)x)/(sqrt(1-x)(sqrt(pi)+sqrt(2 sin^(-1)x))) = lim_(x rarr 1^(-))(1)/(2sqrt(pi)) xx (pi - 2 sin^(-1)x)/(sqrt(1-x))`
Put `x = sin theta` then `theta rarr (pi^(-))/(2)` as `x rarr 1^(-)`
`= lim_(theta rarr (pi^(-))/(2))(1)/(2sqrt(pi)) xx (pi - 2 sin^(-1)(sin theta))/(sqrt(1- sin theta)) = lim_(theta-(pi^(-))/(2))(1)/(2sqrt(pi)) xx (pi - 2 theta)/(sqrt(1-sin theta))`
`theta = (pi)/(2) - h, h rarr 0 = lim_(h rarr 0)(1)/(2sqrt(pi)) xx (pi - 2((pi)/(2)-h))/(sqrt(1-cos h)) = (1)/(2sqrt(pi)) xx (2h)/(sqrt(2)sin.(h)/(2)) = sqrt((2)/(pi))`