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Question Number 123874 by john_santu last updated on 29/Nov/20
(1)limθ→0θ−sinθcosθtanθ−θ(2)limx→0sin3x−3x+x2sinxsin2x
Answered by liberty last updated on 29/Nov/20
solvethisquestion(1)limθ→0θ−sinθcosθtanθ−θ(2)limx→0sin3x−3x+x2sinxsin2xSolution:(1)byL′Hopitalrulelimθ→0θ−12sin2θtanθ−θ=limθ→01−cos2θsec2θ−1limθ→0cos2θ(2sin2(θ2))sin2θ=limθ→02(sinθ2tanθ)2=12(2)limx→0sin3x−3x+x2−12(cos3x−cosx)=−2×limx→03cos3x−3+2x−3sin3x+sinx=−2×limx→0−9sin3x+2−9cos3x+cosx=−2×[2−8]=12.
Answered by Dwaipayan Shikari last updated on 29/Nov/20
1)limx→0x−12sin2xx+x33−x=x−x+812x3x33=2(sinx=x−x36)2)limx→03sinx−4sin3x−3x+x2x.2x=3x−4x3−3x+x22x2=12−2x=12(sinx→x)
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