Question #dc296

1 Answer
Cesareo R.
Nov 24, 2016

See below.

Explanation:

#(cos^2x-cos^2y)/(x^2-y^2)=((cosx+cosy)/(x+y))((cosx-cosy)/(x-y))# so
#lim_(x->y)(cos^2x-cos^2y)/(x^2-y^2)=lim_(x->y)((cosx+cosy)/(x+y))lim_(x->y)((cosx-cosy)/(x-y)) = (2cosy)/(2y)(-siny)=-(cosy siny)/y#

but #sin(2y)=2cosy siny# so finally

#-(cosy siny)/y=-sin(2y)/(2y)#

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