lim(x→0)((1+x)^1/2)-1/sinx
=lim(x→0)[(1+x)^(1/2)-1]'/(sinx)'
=lim(x→0){1/2[(1+x)^(1/2)]}/cosx
=[1/2(1+0)^(1/2)]/cos0
=1/2.