Limit sin2x/sin3x
x>0

see picture

Step-by-step explanation:

Explanation:

Since we have an indeterminate form of type 00, we can apply the l'Hopital's rule:

limx→0  sin(2x)/sin(3x)=  limx→0  d/dx(sin(2x))/ d/dx(sin(3x))

limx→0  d/dx(sin(2x))/d/dx(sin(3x) ) = limx→02cos(2x)/3cos(3x)

Substitute the variable with the value:

limx→02cos(2x)3cos(3x)=(23)

Therefore,

limx→0sin(2x)sin(3x)=23

limx→0sin(2x)sin(3x)=2/3

i'm sure goog can you get that answer

i believe the correct answer is b.