Limit as x goes to infinity (sin3x)/(sin2x)

can you please explain me how it works? and dont use lhopilals rule, cause this is something other..

tnx

lim x→∞ sin(3x)/sin(2x)
lim x→∞ (sin(2x + x))/sin(2x)

Use some identities:
sin(α + ß) = sin(α)cos(ß) + cos(α)sin(ß)
sin(2x + x) = sin(2x)cosx + cos(2x)sinx

sin(2x) = 2sinxcosx
cos(2x) = cos²x - sin²x

lim x→∞ [sin(2x)cosx + cos(2x)sinx] / [2sinx cosx]
lim x→∞ [2sinx cosx cosx + (cos²x - sin²x)sinx] / [2sinx cosx]
lim x→∞ [2sinx cos²x + sinx cos²x - sin³x] / [2sinx cosx]
lim x→∞ [3sinx cos²x - sin³x] / [2sinx cosx]
lim x→∞ [3sinx (1 - sin²x) - sin³x] / [2sinx cosx]
lim x→∞ [3sinx - 3sin³x - sin³x] / [2sinx cosx]
lim x→∞ [3sinx - 4sin³x] / [2sinx cosx]
lim x→∞ [3 - 4sin²x] / [2cosx]
lim x→∞ [3 - 4(1 - cos²x)] / [2cosx]
lim x→∞ [-1 + 4cos²x] / [2cosx]
lim x→∞ -1/(2cosx) + lim x→∞ 4cos²x/(2cosx)
Since cosine oscillates between -1 and 1 as x approaches infinity, neither limit exists, so the limit of the original function does not exist as x approaches infinity.

- - - - - - - -
lim x→0 sin(3x)/sin(2x)
Using the same steps above, we come to this:
lim x→0 -1/(2cosx) + lim x→0 4cos²x/(2cosx)
- lim x→0 1/(2cosx) + lim x→0 2cosx
-(1/(2(1))) + 2(1)
-(1/2) + 2
3/2

Without using any theorems or rules, you can simply evaluate this limit by figuring out which function travels faster.

since sin3x > sin2x, sin3x travels faster, so you can say that the limit of this function travels to infinity.

Just for fun, if the denominator traveled faster, the entire function would go to 0.

Hope that helps!

sin 3x = 3sinx - 4 (sinx)^3
sin 2x = 2 sinx cosx


f= sin3x/sin2x = (3-4 (sinx)^2)/(2cosx)

at infinity it does not converges to a uniqe value,