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Question Number 101422 by john santu last updated on 02/Jul/20

lim_(h→0 )  ((sin ((α+h)^2 )−sin (α^2 ))/(cos ((α+h)^2 sin (α+h)−cos (α^2 )sin (α))) =?

limh0sin((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

⇔((lim_(h→0) ((sin ((α+h)^2 −sin (α^2 ))/h))/(lim_(h→0) ((cos ((α+h)^2 sin (α+h)−cos (α^2 )sin (α))/h))) =  let f(α) = sin (α^2 ) →f ′(α)= 2α cos (α^2 )  let g(α) = cos (α^2 )sin (α)→g ′(α)=−2αsin (α^2 )sin (α)+cos (α^2 )cos (α)  therefore we get limit  = ((2α cos (α^2 ))/(−2α sin (α^2 )sin (α)+cos (α^2 )cos (α))) ★

limh0sin((α+h)2sin(α2)hlimh0cos((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

lim_(h→0) ((sin((α+h)^2 )−sin(α^2 ))/h).(h/(cos((α+h)^2 )sin(α+h)−cosα^2 sinα))                       ((2αcosα^2 )/(−2αsinα^2 sinα+cosαcosα^2 ))=((2α)/(−2αtanα^2 sinα+cosα))★■L

limh0sin((α+h)2)sin(α2)h.hcos((α+h)2)sin(α+h)cosα2sinα2αcosα22αsinα2sinα+cosαcosα2=2α2αtanα2sinα+cosαL

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