# Math Help - Trig limit help

1. ## Trig limit help

lim x -> 0 of ( (2 - cos(3x) - cos(4x) ) / x )

Is there some identity that I can use that I'm not seeing.. ? I can't multiply by a conjugate can I?

2. Originally Posted by nautica17

lim x -> 0 of ( (2 - cos(3x) - cos(4x) ) / x )

Is there some identity that I can use that I'm not seeing.. ? I can't multiply by a conjugate can I?
The slacker's way is to use l'Hopital's Rule.

Otherwise you could substitute the Maclaurin series for cos(3x) and cos(4x), simplify the numerator and then cancel the common factor of x. Then take the limit.

Alternatively, you could probably do a clever re-arrangement and get standard forms whose limits are well known. But why hoe the hard road?

3. The simple little fact that $\frac{1-\cos x}x\to0$ as $x\to0$ will solve this:

$\frac{2-\cos 3x-\cos 4x}{x}=3\cdot \frac{1-\cos 3x}{3x}+4\cdot \frac{1-\cos 4x}{4x}.$ So the limit is zero.

4. Originally Posted by Krizalid
The simple little fact that $\frac{1-\cos x}x\to0$ as $x\to0$ will solve this:

$\frac{2-\cos 3x-\cos 4x}{x}=3\cdot \frac{1-\cos 3x}{3x}+4\cdot \frac{1-\cos 4x}{4x}.$ So the limit is zero.
That is an awesome trick! I give you kudos for that one. It's amazing how there so many ways to solve a math problem.

5. Originally Posted by nautica17

lim x -> 0 of ( (2 - cos(3x) - cos(4x) ) / x )

Is there some identity that I can use that I'm not seeing.. ? I can't multiply by a conjugate can I?
If you were going to be a slacker (as I am ), you would notice that the limit tends to $\frac{0}{0}$ by direct substitution.

So you can use L'Hospital's Rule.

$\lim_{x \to 0}\frac{2 - \cos{3x} - \cos{4x}}{x} = \lim_{x \to 0}\frac{\frac{d}{dx}(2 - \cos{3x} - \cos{4x})}{\frac{d}{dx}(x)}$

$= \lim_{x \to 0}\frac{3\sin{3x} + 4\sin{4x}}{1}$

$= \frac{3\cdot 0 + 4\cdot 0}{1}$

$= 0$, as shown previously.