limx→0 (1−cos2x)22xtanx−xtan2x, is
2
−12
12
-2
limx→0 (1−cos2x)22xtanx−xtan2x
=limx→0 4sin4xx2tanx−2tanx1−tan2x=limx→0 4sin4x1−tan2x−2xtan3x=−2limx→0 sinxxcos3x1−tan2x=−2×1×1=−2