limx→1−√π−√2sin−1x√1−x is equal to

limx→1−√π−√2sin−1x√1−x × √π+√2sin−1x√π+√2sin−1x
limx→1−(π−2sin−1x)√1−x(√π+√2sin−1x)
limx→1−−cos−1x√1−x × 1/(2√π)
Put x = cosθ
limθ→0+2θ√2sin(θ/2) × 1/(2√π) = √2π