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Question:

limx→0xtan2x𕒶xtanx(1−cos2x)2equals.112𕒵314limx→0xtan2x𕒶xtanx(1−cos2x)2equals.112𕒵314limx→0xtan2x𕒶xtanx(1−cos2x)2limx→0xtan2x𕒶xtanx(1−cos2x)2limx→0xtan2x𕒶xtanx(1−cos2x)2limx→0xtan2x𕒶xtanx(1−cos2x)2limx→0xtan2x𕒶xtanx(1−cos2x)2limx→0limx→0limlimlimlimlimx→0x→0x→0x→0xx→→00xtan2x𕒶xtanx(1−cos2x)2xtan2x𕒶xtanx(1−cos2x)2xtan2x𕒶xtanx(1−cos2x)2xtan2x𕒶xtanx(1−cos2x)2xtan2x𕒶xtanxxtan2x𕒶xtanxxxtantan22xx−𕒶2xxtantanxx(1−cos2x)2(1−cos2x)2((11−−coscos22xx)2)))222221111111212121212121212111222𕒵3𕒵3𕒵3𕒵3−𕒵31313131113331414141414141414111444A𕒵3𕒵3𕒵3𕒵3𕒵3−𕒵3131313111333B1111111C121212121212121212111222D141414141414141414111444?

𕒵3

1

12

14

Solution:

Given limit isL=limx→0(xtan2x𕒶xtanx)(1−cos2x)2By expandingtan2xandcos2xwe get(xtan2x𕒶xtanx)(1−cos2x)2=x2tanx1−(tanx)2𕒶xtanx(1−(1𕒶sin2x))2=2xtanx−[2xtanx𕒶xtan3x]4sin4x×(1−tan2x)=2xtan3x4sin4x×(1−tan2x)=2xtan3x4sin4x×(cos2x−sin2xcos2x)=2xsin3xcos3x4sin4x×(cos2x−sin2xcos2x)=x2sinx×(cos2x−sin2x)cosxNow applying the limitx→0,we getL=limx→0x2sinx×limx→01cosx(cos2x−sin2x)=limx→0x2sinx×limx→01cos0(cos20−sin20)=12[∵limx→0xsinx=1]Hence, option D is correct.