# Homework Help: Sigh more limit problems

1. Sep 25, 2007

### physstudent1

1. The problem statement, all variables and given/known data

Sadly I am having more limit problems. This one involving trig.

$\lim_{x \rightarrow 0} \frac{cos2x-cosx}{x}$

2. Relevant equations

3. The attempt at a solution

I am trying to work to get to terms of (1-cosx)/x because that is 0, however I am getting stuck over and over again I keep trying to do things with trig identities and double angle formulas but I am getting nowhere the answer on the calculator is 0, but I can't get to that on paper. I broke the cos2x into 2cos^2 x -1 so then I had (2cos^2x-1-cosx )/x
breaking these up into

$\lim_{x \rightarrow a} \frac{(-1-cos2x)}{x}$-$\lim_{x \rightarrow a} \frac{(1-cosx)}{x}$

the 2nd limit will be 0 but what can I do with the first.

2. Sep 25, 2007

is that (cosx)^2 or cos(2x)?

3. Sep 25, 2007

### physstudent1

cos(2x) after using the double angle formula for cos2x I used the cos^2x one to get what is up there

4. Sep 25, 2007

### anthonym44

haha i was just going to post this same problem

5. Sep 26, 2007

### physstudent1

anyone

6. Sep 26, 2007

### ZioX

This is obviously a L'Hopital problem.

7. Sep 26, 2007

### physstudent1

no we didn't learn L'hopital's rule yet in this class

8. Sep 26, 2007

### Hurkyl

Staff Emeritus
That first one doesn't exist, but that's because you made a sign error.

9. Sep 26, 2007

### Avodyne

Yes, fix the sign error. Then, if the limit of (1-cos x)/x is zero, what is the limit of (1-cos 2x)/(2x) ?

10. Sep 26, 2007

### physstudent1

I reevaluated and I got (1+cos2x)/x for the first part but this isn't (1-cos2x)/(2x)

I had 2cos^2x so I got 2((1+cos2x)/2)/x = (1+cos2x)/(x)

11. Sep 26, 2007

### kidwithshirt

hospital time!

turns out to be lim x->0 (2sin(2x)+sinx)/1

have fun

12. Sep 26, 2007

### physstudent1

what? if your using hopital's I can't use that we didn't get there yet otherwise how did you get to that I am confused.

13. Sep 26, 2007

### physstudent1

I'm not sure where to go from (1+cos2x)/(x) for the first limit I tried to make it to sin making cos2x=2sin^2x so i had 1+2sin^x then 1-2(1-cos^x) / x which turnd to -1(1-cos^2x)/x for the first limit can someone help

14. Sep 26, 2007

### Avodyne

You still have a sign error. The formula you want to use is

cos(2x)-cos(x) = [1-cos(x)]-[1-cos(2x)]

Or, since you're interested in the x->0 limit, you could Taylor expand each cosine: cos(x)=1-(1/2)x2+...

Last edited: Sep 26, 2007
15. Sep 26, 2007

### Johan de Vries

In the numerator of the first limit you made a sign error, the minus 1 should be 1. If you substitute x = y/2 there, then it becomes very similar to the second limit :)

16. Sep 26, 2007

### physstudent1

i don't see how I made the sign error though I used cos(2x) = 2cos^2x-1 so

2cos^2x-1 - cosx is correct so far right?

then I split it into two limits

2cos^2x - 1-cos(x)

then (2(1+cos(2x))/2)/x - 1-cos(x) ?

17. Sep 26, 2007

### Avodyne

I can't follow what you're doing. You need to write whole equations, with equal signs.

But you're making this much more complicated than it needs to be, as should be clear from Johan's post.

18. Sep 27, 2007

### physstudent1

I understand it now thanks guys.