lim [sin(2x)/sin(3x) + sin(3x)/4x]
x->0

kthx

if sin 3x = 3 sin x – 4 sin^3 ( x ), and sin2x = 2sinxcosx, then

[(2sinxcosx )/ (3sinx - 4 sin^3( x ))] + [3sinx - 4 sin^3( x ) / 4x] =

(sinx/sinx)[2cosx / 3 - 4 sin^2 ( x )] + [3sinx/4x - 4sin^3( x )/4x]=

[2cosx / 3 - 4 sin^2( x )] + 3/4(sinx/x) - sin^3( x )/x =

lim x-> 0 = 2/3 + 3/4 - 1 = 5/12

Don’t take my word for it though, I probably did a mistake. I haven’t done math like this in so long.

oh my god. so much sin

I saw the sin; it opened up my eyes! I saw the sin!

From my talk with Klez, I have part of it right, part of it wrong. I think I have the limit of sin^3 (x) / x wrong, I had to make an assumption on that part.

Thanks for getting that stuck in my head

So what…level of math is this? Basic high school math, or ultra high tech university math?

looks more like chemistry to me… atleast the sin part does.

Calculus?

This is a limit problem, this is the preliminary method to derivatives (one of the main calculus components). Its a mean limit problem if you ask me. Basic calculus is really not much more than 2 equations: 1 for the derivative, 1 for the integral. Everything else is learning how to do algebra tricks or identities to plug equations in either of the 2 calculus methods.

Limits of trig functions have always given me headaches… It looks like something to use L’Hopital’s rule on and perhaps split into 2 separate limits, but I really don’t want to think about it any further than that.

Actually I just remembered a trick where

lim x->0 [sin(2x)/sin(3x) + sin(3x)/4x]

lim x->0 [{sin(2x)/x} / {sin(3x)/x} ] =

(2/3) lim x->0 [{sin(2x)/2x} / {sin(3x)/3x} ] = 2/3.

The question I have is am I allowed to break that limit into 2 parts and do the same to sin(3x)/4x? Then it would be (3/4) lim x->0 = sin (3x) / 3x which would be much friendlier. 2/3 + 3/4 = 17/12. I’m not sure if that’s legit though. It would certainly invalidate what I wrote in my first reply to Klez. It would make this problem a whole lot easier though.

I believe you are, but my calculus experience is limited to an issue with trigonometry proofs I took waaaaaay too far with my teacher, and it got into limits.

What the heck are they teaching you guys? I guess Sin’s crazy methods work if you’re not supposed to use l’hopital’s rule. But if you can use l’hopital’s rule, it makes these really easy. I guess if you can’t use it, Sin’s newest method works. Assuming you can use lim x->0 sin(x)/x = 1.

You are allowed to break it into two limits. This can be done as long as the limits of both parts actually exist and the result isn’t an indeterminate form…

Split it into two parts:

lim [sin(2x)/sin(3x)]
x -> 0

lim [sin(3x)/4x]
x -> 0

Use L’Hopital’s rule on both parts to get the equivalent statement:

lim [2cos(2x)/3cos(3x)]
x -> 0

lim [3cos(3x)/4]
x -> 0