Question #a1349

1 Answer
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