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Question Number 139001 by bramlexs22 last updated on 21/Apr/21
limx→01+sin2x3−1−2tanx4sinx+tan2x=?
Answered by EDWIN88 last updated on 21/Apr/21
limx→0cos2x(1+sin2x3−1−2tanx4cos2xsinx+sin2x)=limx→0(1+x23)−(1−2x4)x(1+x)=limx→0x(x3+12)x(1+x)=12
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