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Limit sin2x/sin3x
x>0

User Jekcom
by
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1 Answer

1 vote

Answer:

see picture

Step-by-step explanation:

Step-by-step 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

Answer: limx→0sin(2x)sin(3x)=2/3

Limit sin2x/sin3x x>0-example-1
User Vincent Gigandet
by
4.9k points