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Question Number 117422 by bemath last updated on 11/Oct/20
limx→03tan4x−4tan3x3sin4x−4sin3x=?
Answered by bobhans last updated on 11/Oct/20
limx→03tan4x−4tan3x3sin4x−4sin3x=limx→03(4x+13(4x)3)−4(3x+13(3x)3)3(4x−(4x)36)−4(3x−(3x)36)=limx→012x+64x3−12x−36x312x−32x3−12x+18x3=limx→028x3−14x3=−2
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