lim(x→0) (x-arcsinx)/(sinx)^3的解题过程通过L'Hopital's Rule （洛必达法则）来解答

lim[x→0] (x-arcsinx)/sin³x
=lim[x→0] [1-1/√(1-x²)]/(3sin²xcosx)
=(1/3)lim[x→0] [-x/(1-x²)^(3/2)]/(2sinxcos²x-sin³x)
=(-1/6)lim[x→0] [(2x²+1)/(1-x²)^(5/2)]/(cos³x-5sin²xcosx)
=(-1/6)*[(0+1)/(1-0)^(5/2)]/(1-5*0)
=-1/6