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))) =?

Commented byjohn 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))) ★

Commented byDwaipayan 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