what is the limit of sin2x/sin3x as x approaches 0?

Question

i got 2/3 first but i think it might be 6

Answer 1

Use L'Hopital's rule take derivative of sin(2x) and derivative of sin(3x)

d(sin(2x))/dx = 2cos(2x) and d(sin(3x))/dx = 3cos(3x)

Now take limit of 2cos(2x)/(3cos(3x)) as x--> 0 you get 2/3.

Answer 2

sin2x/sin3x
= (sin(2x))/(2x) * (3x)/sin(3x) * (2/3)
As x approches zero
(sin(2x))/2x approaches 1
((3x)/sin(3x) approaches 1

The required limit is 2/3

You can multiply by x/x because lim(x->0) [x/x] = 1.

Answer 3

You can use the L'Hopital's rule take derivative of sin(2x) and derivative of sin(3x)
remember, you cannot multiply and divide by 'x' as 'x' tends to zero.

d(sin(2x))/dx = 2cos(2x)

d(sin(3x))/dx = 3cos(3x)

taking limit of 2cos(2x)/(3cos(3x)) as x tends to 0 you get

2/3

Answer 4

zero