The correct option is D 8
limx→π4cot3x−tanxcos(x+π4)=limx→π41−tan4xtan3x(1√2cosx−1√2sinx)=limx→π4(1+tan2x)(1−tan2x)1√2tan3x.(cosx−sinx)=√2limx→π4(1+1).(cos2x−sin2x).cos3xsin3x.cos2x.(cosx−sinx)=2√2limx→π4(cosx+sinx)(cosx−sinx).cosxsin3x.(cosx−sinx)=2√2.(1√2+1√2)⋅1√2(1√2)3=2√2⋅2√2⋅1=8