Question

Question #a1349

Answer 1

`-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`