Question #a1349

1 Answer
Cesareo R.
Apr 6, 2017

#-8#

Explanation:

Applying l'Hopital's rule

#lim_(x->0)(-5sin(5x)+3sin(3x))/(2x)=1/2lim_(x->0)(-5sin(5x)/x+3sin(3x)/x)=#
#=1/2lim_(x->0)(-5^2sin(5x)/(5x)+3^2sin(3x)/(3x))#

but #lim_(y->0)sin(y)/y = 1#

so

#lim_(x->0)(cos(5x)-cos(3x))/x^2 =#

#=1/2lim_(x->0)(-5^2sin(5x)/(5x)+3^2sin(3x)/(3x))=1/2(3^2-5^2)=-8#

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