limx→2[1x(x−2)2−1(x2−3x+2)]
=limx→2[1x(x−2)2−1(x−1)(x−2)]
=limx→2[x−1−x(x−2)x(x−2)2]
=limx→2[x−1−x2+2xx(x−2)2]
=limx→2[3x−x2−1x(x−2)2]=limx→2−(x2−3x+1)x(x−2)2
limx→2−(x2−3x+1)x(x−2)2
=−(4−6+1)2×0
=−(5−6)0
=∞
limx→π22−cosx−1x(x−π2)