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Question Number 101422 by john santu last updated on 02/Jul/20
limh→0sin((α+h)2)−sin(α2)cos((α+h)2sin(α+h)−cos(α2)sin(α)=?
Commented by john santu last updated on 02/Jul/20
⇔limh→0sin((α+h)2−sin(α2)hlimh→0cos((α+h)2sin(α+h)−cos(α2)sin(α)h=letf(α)=sin(α2)→f′(α)=2αcos(α2)letg(α)=cos(α2)sin(α)→g′(α)=−2αsin(α2)sin(α)+cos(α2)cos(α)thereforewegetlimit=2αcos(α2)−2αsin(α2)sin(α)+cos(α2)cos(α)★
Commented by Dwaipayan Shikari last updated on 02/Jul/20
limh→0sin((α+h)2)−sin(α2)h.hcos((α+h)2)sin(α+h)−cosα2sinα2αcosα2−2αsinα2sinα+cosαcosα2=2α−2αtanα2sinα+cosα★◼L
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