Struggling with Trig Limits: How to Simplify cos2x-cosx/x?

[physstudent1]

Homework Statement

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

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

Homework Equations

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.

 

[silvashadow]

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

 

[physstudent1]

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

 

[anthonym44]

haha i was just going to post this same problem

 

[physstudent1]

anyone

 

[ZioX]

This is obviously a L'Hopital problem.

 

[physstudent1]

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

 

[Hurkyl]

I am trying to work to get to terms of (1-cosx)/x because that is 0
...
\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.

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

 

[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) ?

 

[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)

 

[kidwithshirt]

hospital time!

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

have fun

 

[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.

 

[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

 

[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+...

 

[Johan de Vries]

Homework Statement

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.

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 :)

 

[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) ?

 

[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.

 

[physstudent1]

I understand it now thanks guys.