Solving Tricky Limits Using Trigonometric Identities

[The_Brain]

Hey, I have this limit question that I just cannot find the trick for. I've even tried using multiple-angle formulas and converting to complex numbers but I just can't seem to get it. The limit is:

(2sin(x)-sin(2x))/(3sin(x)-sin(3x)) as x -> 0.

The supposed answer is 1/4. Any insights are greatly appreciated.

 

[Physics Monkey]

Its and indeterminate form, try L'Hospital's Rule. Hint: twice.

 

[AKG]

Show your work.

 

[AKG]

As far as I can tell, L'Hopital's rule is a terrible way to go about doing it. It can be solved very quickly with trig identities.

 

[courtrigrad]

Couldn't you simplify it by doing: \lim_{x\rightarrow 0} \frac{2\sin x - \sin 2x}{3\sin x - \sin 3x} = \frac{2}{3}(\frac{\sin x - \sin 2x}{\sin x - \sin 3x})? Since you know the answer is \frac{1}{4} you the the expression in the parantheses has to be \frac{3}{8}.

Just an idea

Maybe you could rationalize the denominator

 

[The_Brain]

I figured it out! Use trig identities for sin(2x) and sin(3x) as well as a double L'Hospital's. In all honesty, I really should have done that before giving up and posting here; I really appreciate all the quick replies, though!

Edit: Actually it's 3 applications of L'Hospitals.

 

[AKG]

I take back what I said. I'm not sure that L'Hoptial's rule is a bad way to go. I don't know if it is any good, but I initially thought it was bad because I wasn't thinking (I was thinking you'd have to find the derivative of the whole expression, and you'd get ugly stuff from the quotient rule, but then I remembered that you just have to differentiate top and bottom separately). Anyways, using trig identities (once for the top, twice for the bottom) the problem can be solved nicely and neatly in about 7-8 lines of work, so like I said, I don't know whether L'Hoptial's is good or bad, but if it is more than 10-12 lines, it's not the best way to go.